129
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$x,y,z >0$, prove $$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$

Note: Often Stack Exchange asked to show some work before answering the question. This inequality was used as a proposal problem for National TST of an Asian country a few years back. However, upon receiving the official solution, the committee decided to drop this problem immediately. They don't believe that any students can solve this problem in 3 hour time frame.

Update 1: In this forum, somebody said that BW is the only solution for this problem, which to the best of my knowledge is wrong. This problem is listed as "coffin problems" in my country. The official solution is very elementary and elegant.

Update 2: Although there are some solutions (or partial solution) based on numerical method, I am more interested in the approach with "pencil and papers." I think the approach by Peter Scholze in here may help.

Update 3: Michael has tried to apply Peter Scholze's method but not found the solution yet.

Update 4: Symbolic expanding with computer is employed and verify the inequality. However, detail solution that not involved computer has not been found. Whoever can solve this inequality using high school math knowledge will be considered as the "King of Inequality".

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    $\begingroup$ I have no idea about the official solution. I try this problem for the past 3 years but not yet success. Even with brute force, I still cannot solve it. This shows level of insanity this problem has. $\endgroup$
    – HN_NH
    May 7, 2016 at 16:37
  • 9
    $\begingroup$ A simple observation is that the inequality is homogeneous, so it suffices to prove the case $ x + y + z = 1 $. Wolfram helped me solve the mess of an equation system that arises out of Lagrange multipliers, so I am convinced that the inequality is true now. (I have no idea how to solve the system myself, so this doesn't really count as a solution.) $\endgroup$
    – Ege Erdil
    May 7, 2016 at 23:01
  • 2
    $\begingroup$ I believe that $\sum_{\mathrm{cyc}}\frac{x^4}{ay^3+bz^3}\geq\frac{x+y+z}{a+b}$, and why not $\sum_{\mathrm{cyc}}\frac{x_n^4}{a_1x_1^3+\ldots a_{n-1}x_{n-1}^3}\geq\frac{x_1+\ldots+x_{n-1}+x_n}{a_1+\ldots+a_{n-1}+a_n}$ for $n\geq3$. Maybe the general case could be somehow helpful... $\endgroup$
    – Nicolas
    May 22, 2016 at 9:13
  • 3
    $\begingroup$ Any further ideas for changes of the tags here must be first proposed to me. $\endgroup$
    – quid
    Oct 13, 2018 at 23:22
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    $\begingroup$ @YuriNegometyanov: If you have a solution then why don't you just post it? – I find this request for a bounty a bit strange, and to be honest, I wonder that a moderator complied with it. (It isn't just for the hat, or is it?) $\endgroup$
    – Martin R
    Dec 29, 2019 at 13:23

15 Answers 15

31
$\begingroup$

A big problem we get around $(x,y,z)=(0.822,1.265,1.855)$.

The Buffalo Way helps:

Let $x=\min\{x,y,z\}$, $y=x+u$,$z=x+v$ and $x=t\sqrt{uv}$.

Hence, $\frac{13}{5}\prod\limits_{cyc}(8x^3+5y^3)\left(\sum\limits_{cyc}\frac{x^4}{8x^3+5y^3}-\frac{x+y+z}{13}\right)=$

$$=156(u^2-uv+v^2)x^8+6(65u^3+189u^2v-176uv^2+65v^3)x^7+$$ $$+2(377u^4+1206u^3v+585u^2v^2-1349uv^3+377v^4)x^6+$$ $$+3(247u^5+999u^4v+1168u^3v^2-472u^2v^3-726uv^4+247)x^5+$$ $$+3(117u^6+696u^5v+1479u^4v^2+182u^3v^3-686u^2v^4-163uv^5+117v^6)x^4+$$ $$+(65u^7+768u^6v+2808u^5v^2+2079u^4v^3-1286u^3v^4-585u^2v^5+181uv^6+65v^7)x^3+$$$$+3uv(40u^6+296u^5v+472u^4v^2-225u^2v^4+55uv^5+25v^6)x^2+ $$ $$+u^2v^2(120u^5+376u^4v+240u^3v^2-240u^2v^3-25uv^4+75v^5)x+$$ $$+5u^3v^3(8u^4+8u^3v-8uv^3+5v^4)\geq$$ $$\geq u^5v^5(156t^8+531t^7+2t^6-632t^5-152t^4+867t^3+834t^2+299t+40)\geq0$$

Done!

For example, we'll prove that $$6(65u^3+189u^2v-176uv^2+65v^3)\geq531\sqrt{u^3v^3},$$ which gives a coefficient $531$ before $t^7$ in the polynomial $156t^8+531t^7+2t^6-632t^5-152t^4+867t^3+834t^2+299t+40.$

Indeed, let $u=k^2v$, where $k>0$.

Thus, we need to prove that: $$130k^6+378k^4-177k^3-352k^2+130\geq0$$ and by AM-GM we obtain: $$130k^6+378k^4-177k^3-352k^2+130=$$ $$=130\left(k^3+\frac{10}{13}k-1\right)^2+\frac{k}{13}(2314k^3+1079k^2-5576k+2600)\geq$$ $$\geq\frac{k}{13}\left(8\cdot\frac{1157}{4}k^3+5\cdot\frac{1079}{5}k^2+21\cdot\frac{2600}{21}-5576k\right)\geq$$ $$\geq\frac{k^2}{13}\left(34\sqrt[34]{\left(\frac{1157}{4}\right)^8\left(\frac{1079}{5}\right)^5\left(\frac{2600}{21}\right)^{21}}-5576\right)>0.$$ We'll prove that $$ 2(377u^4+1206u^3v+585u^2v^2-1349uv^3+377v^4)\geq2u^2v^2,$$ for which it's enough to prove that: $$377t^4+1206t^3+584t^2-1349t+377\geq0$$ or $$t^4+\frac{1206}{377}t^3+\frac{584}{377}t^2-\frac{1349}{377}t+1\geq0$$ or $$\left(t^2+\frac{603}{377}t-\frac{28}{29}\right)^2+\frac{131015t^2-69589t+9633}{142129}\geq0,$$ which is true because $$69589^2-4\cdot131015\cdot9633<0.$$

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    $\begingroup$ It seems that everybody overlooked that this is the solutions of this problem. I changed the last step of the proof so hat it is simpler. I hope thi is ok. I also added an answer that is an extenede comment to this proof the shows how to confirm these calculations with the cas Maxima. $\endgroup$
    – miracle173
    Jul 3, 2016 at 12:16
  • 1
    $\begingroup$ there is an error in my edit. One cannot assume $z\le y$. I try do undo the change. $\endgroup$
    – miracle173
    Jul 3, 2016 at 13:06
  • 1
    $\begingroup$ @MichaelRozenberg: can you elaborate how you get the the polynomial in t and the estimates? $\endgroup$
    – miracle173
    Jul 3, 2016 at 13:22
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    $\begingroup$ @miracle173 for example, $156(u^2-uv+v^2)x^8\geq156uvx^8=156u^5v^5t^8$. $\endgroup$ Jul 3, 2016 at 13:31
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    $\begingroup$ Here is a link to an answer that contains a Maxima script with some details to this answer. $\endgroup$
    – miracle173
    Jul 3, 2016 at 18:30
25
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enter image description here

This is a question of the symmetric type, such as listed in:

With a constraint $\;x+y+z=1\;$ and $\;x,y,z > 0$ . Sort of a general method to transform such a constraint into the inside of a triangle in 2-D has been explained at length in:

Our function $f$ in this case is: $$ f(x,y,z) = \frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} - \frac{1}{13} $$ And the minimum of that function inside the abovementioned triangle must shown to be greater or equal to zero. Due to symmetry - why oh why can it not be proved with Group Theory - an absolute minimum of the function is expected at $(x,y,z) = (1/3,1/3,1/3)$. Another proof without words is attempted by plotting a contour map of the function, as depicted. Levels (nivo) of these isolines are defined (in Delphi Pascal) as:

nivo := min + sqr(g/grens)*(max-min); { sqr = square ; grens = 20 ; g = 0..grens }
The whiteness of the isolines is proportional to the (positive) function values; they are almost black near the minimum and almost white near the maximum values. Maximum and minimum values of the function are observed to be:

 0.00000000000000E+0000 < f < 4.80709198767699E-0002
The little $\color{blue}{\mbox{blue}}$ spot in the middle is where $\,0 \le f(x,y,z) < 0.00002$ .

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    $\begingroup$ Interesting technique. Thank you very much for trying my inequality. One up vote !!! $\endgroup$
    – HN_NH
    May 24, 2016 at 6:05
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+25
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Too long for a comment.


The Engel form of Cauchy-Schwarz is not the right way:

$$\frac{(x^2)^2}{8x^3+5y^3}+\frac{(y^2)^2}{8y^3+5z^3}+\frac{(z^2)^2}{8z^3+5x^3} \geq \frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}$$

So we should prove that $$\frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}\geq\frac{x+y+z}{13}$$

which is equivalent to $$\frac{(x^2+y^2+z^2)^2}{(x^3+y^3+z^3)}\geq x+y+z$$ but by Cauchy-Schwarz again we have $$x+y+z=\frac{(x^2)^2}{x^3} +\frac{(y^2)^2}{y^3}+\frac{(z^2)^2}{z^3} \geq \frac{(x^2+y^2+z^2)^2}{(x^3+y^3+z^3)}$$

and the inequalities are in the wrong way.

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    $\begingroup$ So directly using CS is too weak. Any other stronger inequalities that we can try? $\endgroup$
    – Yuxiao Xie
    May 22, 2016 at 5:56
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    $\begingroup$ I've been asked to delete my answer, and I want to explain why I posted it. I know that this is not an answer to the original problem. I posted it because there where another anwer claiming that the problem could have been solved by directly applying the Cauchy Schwarz inequality. So I though that was a good idea to show that this way is not the right way. Now the other answer has been deleted, but I still think that it can be useful to note that a direct application of CS does not work. $\endgroup$
    – user126154
    Jun 23, 2016 at 9:03
  • $\begingroup$ This is very interesting... instead of an answer showing what works, we have an answer showing what doesn't work, which could prove to be equally useful! Love it! $(+1)$ :D $\endgroup$
    – Mr Pie
    Feb 2, 2019 at 13:58
8
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This is more an extended comment to the answer of @MichaelRozenberg than an answer by its own. I used a short Maxima to confirm the equation derived by @MichaelRozenberg. I used Maxima because it is open source.

Here is the Maxima script (statements are terminated by $ or by ;):


"I use string to comment this file"$

"the flag `display2d`  controls 
the display of the output. You can unset it (display2d:false), that makes it easy to copy 
the maxima output to math.stackexchange"$

"to make it easier to input the problem data 
we define to function g and f:"$

g(r,s):=(8*r^3+5*s^3);

f(r,s):=r^4/g(r,s);

"
the initial problem has the form 
L(x,y,t)>=R(x,y,z) 
but we subtract R(x,y,z) from this equation and 
we state the problem in the form 
term0>=0 
where term0 is L(x,y,z)-R(x,y,z) 
this is term0:
"$

term0:f(x,y)+f(y,z)+f(z,x)-(x+y+z)/13;

"
Now we multiply the term0 by a positive fraction of the (positive) common denominator 
and get term1 that satisfies 
term1>=0 
`ratsimp` does some simplification like cancelling 
"$

term1:13/5*g(x,y)*g(y,z)*g(z,x)*term0,ratsimp;

"
now we assume x=0 and v>=0
`,y=x+u` and `,z=x+v` do these substitutions
"$

term2:term1,y=x+u,z=x+v;

"
ratsimp(.,x) does some simplification and displays the term as polynomial of x
"$

term3:ratsimp(term2,x);

for p:0 thru hipow(term3,x) do print (coeff(term3,x,p)*x^p);

"the lowerbound polynomial is given by @Michael Rozenberg";

lowerbound:u^5*v^5*(156*t^8+531*t^7+2*t^6-632*t^5-152*t^4+867*t^3+834*t^2+299*t+40);

"we use the expanded version of the lowerbound polynomial";

lb:lowerbound,expand;

"we want to avoid squareroots and therefore substitute u bei `q^2` and v by `w^2`. 
The expression `sqrt(u*v)` (see thhe proof of Michael Rozenberg) then can be replaced by q*w";

"We want to avoid squareroots and therefore substitute u bei `q^2` and v by `w^2`. 
The expression `sqrt(u*v)` (see thhe proof of Michael Rozenberg) then can be replaced by q*w.
The following loop checks for each exponent k, that the coefficient of the original polynomial 
in x (adjusted by sqrt(u*v)^k) is larger than the coeffiecient of the lowerbound polynomial.
This value is called wdiff in the following.
We already mentioned that we do not use the original variable u and v but first transform 
to q and w as described above and therefor the adjustment is (q*w)^k  instead of sqrt(u*v)^k.
`wdiff` is a homogenous polynom of degree 20. We devide by `w`and replace `q/w` by `s`
and get the polynomial `poly` with vrailbe `s`. For these polynomials we calculate the number
of roots greater than 0. This can be done bei the `nroot` function that uses  'sturm's theorem' 
Then we  calculate the value of poly at 2. If this value is greate 0 and there are 
no zeros greater 0 then wdiff is greater or equal 0 for all nonnegative q and w and therefore 
for all nonegative u and v. This was what we wanted to proof.
We see that all polynomails are positive at 2 and also for all except for k=8 there are no zeros
greater than 0. For k=8 we have a zero with even multiplicity.
";

for k:0 thru 8 do (
    coff_x:coeff(term3,x,k),
    coeff_t:coeff(lb,t,k),
    wdiff:ev(coff_x*(q*w)^k-coeff_t,u=q^2,v=w^2),
    poly:ratsubst(s,q/w,expand(wdiff/w^20)),
    nr:nroots(poly,0,inf),
    print("==="),
    print("k=",k),
    print("coeff(term3, x,",k,")=",coff_x),
    print("coeff(lb, t,",k,")=",coeff_t),
    print("wdiff=",wdiff),
    print("polynomial:",poly),
    print("factors=",factor(poly)),
    print("number of roots >0:",nr),
    print("poly(2)=",ev(poly,s=2))
    );

"finally we proof that the lowerbbound polynomial has no positive root and that 
it is greater than 0 for t=1. Therefor it is greater or equal than 0 for all admissible values";

poly:ratcoeff(lowerbound,u^5*v^5);

poly,t=1;

nroots(poly,0,inf);




I ran the scrip on the Xmaxima console and get the following output. I use this console with this rather ugly kind of output because it can be simply copied and pasted to math.stackecchange. A prettier output can be found here at an online version of Maxima

(%i1) display2d:false;
(%o1) false
(%i2) 
read and interpret file: #pD:/maxima/ineq1775572.mac
(%i3) "I use string to comment this file"
(%i4) "the flag `display2d`  controls 

the display of the output. You can unset it (display2d:false), that makes it easy to copy 

the maxima output to math.stackexchange"
(%i5) "to make it easier to input the problem data 

we define to function g and f:"
(%i6) g(r,s):=8*r^3+5*s^3
(%o6) g(r,s):=8*r^3+5*s^3
(%i7) f(r,s):=r^4/g(r,s)
(%o7) f(r,s):=r^4/g(r,s)
(%i8) "

the initial problem has the form 

L(x,y,t)>=R(x,y,z) 

but we subtract R(x,y,z) from this equation and 

we state the problem in the form 

term0>=0 

where term0 is L(x,y,z)-R(x,y,z) 

this is term0:

"
(%i9) term0:f(x,y)+f(y,z)+f(z,x)+(-(x+y+z))/13
(%o9) z^4/(8*z^3+5*x^3)+y^4/(5*z^3+8*y^3)+((-z)-y-x)/13+x^4/(5*y^3+8*x^3)
(%i10) "

Now we multiply the term0 by a positive fraction of the (positive) common denominator 

and get term1 that satisfies 

term1>=0 

`ratsimp` does some simplification like cancelling 

"
(%i11) ev(term1:(13*g(x,y)*g(y,z)*g(z,x)*term0)/5,ratsimp)
(%o11) (25*y^3+40*x^3)*z^7+((-40*y^4)-40*x*y^3-64*x^3*y+40*x^4)*z^6
                          +(40*y^6+39*x^3*y^3-40*x^6)*z^4
                          +(40*y^7-64*x*y^6+39*x^3*y^4+39*x^4*y^3-40*x^6*y
                                  +25*x^7)
                           *z^3+((-40*x^3*y^6)-64*x^6*y^3)*z+25*x^3*y^7
                          -40*x^4*y^6+40*x^6*y^4+40*x^7*y^3
(%i12) "

now we assume x=0 and v>=0

`,y=x+u` and `,z=x+v` do these substitutions

"
(%i13) ev(term2:term1,y = x+u,z = x+v)
(%o13) (x+v)^3*(40*(x+u)^7-64*x*(x+u)^6+39*x^3*(x+u)^4+39*x^4*(x+u)^3+25*x^7
                          -40*x^6*(x+u))
 +25*x^3*(x+u)^7+(x+v)*((-40*x^3*(x+u)^6)-64*x^6*(x+u)^3)
 +(x+v)^4*(40*(x+u)^6+39*x^3*(x+u)^3-40*x^6)-40*x^4*(x+u)^6+40*x^6*(x+u)^4
 +(x+v)^6*((-40*(x+u)^4)-40*x*(x+u)^3+40*x^4-64*x^3*(x+u))
 +(x+v)^7*(25*(x+u)^3+40*x^3)+40*x^7*(x+u)^3
(%i14) "

ratsimp(.,x) does some simplification and displays the term as polynomial of x

"
(%i15) term3:ratsimp(term2,x)
(%o15) (156*v^2-156*u*v+156*u^2)*x^8+(390*v^3-1056*u*v^2+1134*u^2*v+390*u^3)
                                     *x^7
                                    +(754*v^4-2698*u*v^3+1170*u^2*v^2
                                             +2412*u^3*v+754*u^4)
                                     *x^6
                                    +(741*v^5-2178*u*v^4-1476*u^2*v^3
                                             +3504*u^3*v^2+2997*u^4*v+741*u^5)
                                     *x^5
                                    +(351*v^6-489*u*v^5-2058*u^2*v^4
                                             +546*u^3*v^3+4437*u^4*v^2
                                             +2088*u^5*v+351*u^6)
                                     *x^4
                                    +(65*v^7+181*u*v^6-585*u^2*v^5
                                            -1286*u^3*v^4+2079*u^4*v^3
                                            +2808*u^5*v^2+768*u^6*v+65*u^7)
                                     *x^3
                                    +(75*u*v^7+165*u^2*v^6-675*u^3*v^5
                                              +1416*u^5*v^3+888*u^6*v^2
                                              +120*u^7*v)
                                     *x^2
                                    +(75*u^2*v^7-25*u^3*v^6-240*u^4*v^5
                                                +240*u^5*v^4+376*u^6*v^3
                                                +120*u^7*v^2)
                                     *x+25*u^3*v^7-40*u^4*v^6+40*u^6*v^4
                                    +40*u^7*v^3
(%i16) for p from 0 thru hipow(term3,x) do print(coeff(term3,x,p)*x^p)
25*u^3*v^7-40*u^4*v^6+40*u^6*v^4+40*u^7*v^3 
(75*u^2*v^7-25*u^3*v^6-240*u^4*v^5+240*u^5*v^4+376*u^6*v^3+120*u^7*v^2)*x 
(75*u*v^7+165*u^2*v^6-675*u^3*v^5+1416*u^5*v^3+888*u^6*v^2+120*u^7*v)*x^2 
(65*v^7+181*u*v^6-585*u^2*v^5-1286*u^3*v^4+2079*u^4*v^3+2808*u^5*v^2+768*u^6*v
       +65*u^7)
 *x^3

(351*v^6-489*u*v^5-2058*u^2*v^4+546*u^3*v^3+4437*u^4*v^2+2088*u^5*v+351*u^6)
 *x^4

(741*v^5-2178*u*v^4-1476*u^2*v^3+3504*u^3*v^2+2997*u^4*v+741*u^5)*x^5 
(754*v^4-2698*u*v^3+1170*u^2*v^2+2412*u^3*v+754*u^4)*x^6 
(390*v^3-1056*u*v^2+1134*u^2*v+390*u^3)*x^7 
(156*v^2-156*u*v+156*u^2)*x^8 
(%o16) done
(%i17) "the lowerbound polynomial is given by @Michael Rozenberg"
(%o17) "the lowerbound polynomial is given by @Michael Rozenberg"
(%i18) lowerbound:u^5*v^5
                     *(156*t^8+531*t^7+2*t^6-632*t^5-152*t^4+867*t^3+834*t^2
                              +299*t+40)
(%o18) (156*t^8+531*t^7+2*t^6-632*t^5-152*t^4+867*t^3+834*t^2+299*t+40)*u^5*v
                                                                            ^5
(%i19) "we use the expanded version of the lowerbound polynomial"
(%o19) "we use the expanded version of the lowerbound polynomial"
(%i20) ev(lb:lowerbound,expand)
(%o20) 156*t^8*u^5*v^5+531*t^7*u^5*v^5+2*t^6*u^5*v^5-632*t^5*u^5*v^5
                      -152*t^4*u^5*v^5+867*t^3*u^5*v^5+834*t^2*u^5*v^5
                      +299*t*u^5*v^5+40*u^5*v^5
(%i21) "we want to avoid suareroots and therefore substitute u bei `q^2` and v by `w^2`. 

The expression `sqrt(u*v)` (see thhe proof of Michael Rozenberg) then can be replaced by q*w"
(%o21) "we want to avoid suareroots and therefore substitute u bei `q^2` and v by `w^2`. 

The expression `sqrt(u*v)` (see thhe proof of Michael Rozenberg) then can be replaced by q*w"
(%i22) "We want to avoid suareroots and therefore substitute u bei `q^2` and v by `w^2`. 

The expression `sqrt(u*v)` (see thhe proof of Michael Rozenberg) then can be replaced by q*w.

The following loop checks for each exponent k, that the coefficient of the original polynomial 

in x (adjusted by sqrt(u*v)^k) is larger than the coeffiecient of the lowerbound polynomial.

This value is called wdiff in the following.

We already mentioned that we do not use the original variable u and v but first transform 

to q and w as described above and therefor the adjustment is (q*w)^k  instead of sqrt(u*v)^k.

`wdiff` is a homogenous polynom of degree 20. We devide by `w`and replace `q/w` by `s`

and get the polynomial `poly` with vrailbe `s`. For these polynomials we calculate the number

of roots greater than 0. This can be done bei the `nroot` function that uses  'sturm's theorem' 

Then we  calculate the value of poly at 2. If this value is greate 0 and there are 

no zeros greater 0 then wdiff is greater or equal 0 for all nonnegative q and w and therefore 

for all nonegative u and v. This was what we wanted to proof.

We see that all polynomails are positive at 2 and also for all except for k=8 there are no zeros

greater than 0. For k=8 we have a zero with even multiplicity.

"
(%o22) "We want to avoid suareroots and therefore substitute u bei `q^2` and v by `w^2`. 

The expression `sqrt(u*v)` (see thhe proof of Michael Rozenberg) then can be replaced by q*w.

The following loop checks for each exponent k, that the coefficient of the original polynomial 

in x (adjusted by sqrt(u*v)^k) is larger than the coeffiecient of the lowerbound polynomial.

This value is called wdiff in the following.

We already mentioned that we do not use the original variable u and v but first transform 

to q and w as described above and therefor the adjustment is (q*w)^k  instead of sqrt(u*v)^k.

`wdiff` is a homogenous polynom of degree 20. We devide by `w`and replace `q/w` by `s`

and get the polynomial `poly` with vrailbe `s`. For these polynomials we calculate the number

of roots greater than 0. This can be done bei the `nroot` function that uses  'sturm's theorem' 

Then we  calculate the value of poly at 2. If this value is greate 0 and there are 

no zeros greater 0 then wdiff is greater or equal 0 for all nonnegative q and w and therefore 

for all nonegative u and v. This was what we wanted to proof.

We see that all polynomails are positive at 2 and also for all except for k=8 there are no zeros

greater than 0. For k=8 we have a zero with even multiplicity.

"
(%i23) for k from 0 thru 8 do
           (coff_x:coeff(term3,x,k),coeff_t:coeff(lb,t,k),
            wdiff:ev(coff_x*(q*w)^k-coeff_t,u = q^2,v = w^2),
            poly:ratsubst(s,q/w,expand(wdiff/w^20)),nr:nroots(poly,0,inf),
            print("==="),print("k=",k),print("coeff(term3, x,",k,")=",coff_x),
            print("coeff(lb, t,",k,")=",coeff_t),print("wdiff=",wdiff),
            print("polynomial:",poly),print("factors=",factor(poly)),
            print("number of roots >0:",nr),print("poly(2)=",ev(poly,s = 2)))
=== 
k= 0 
coeff(term3, x, 0 )= 25*u^3*v^7-40*u^4*v^6+40*u^6*v^4+40*u^7*v^3 
coeff(lb, t, 0 )= 40*u^5*v^5 
wdiff= 25*q^6*w^14-40*q^8*w^12-40*q^10*w^10+40*q^12*w^8+40*q^14*w^6 
polynomial: 40*s^14+40*s^12-40*s^10-40*s^8+25*s^6 
factors= 5*s^6*(8*s^8+8*s^6-8*s^4-8*s^2+5) 
number of roots >0: 0 
poly(2)= 769600 
=== 
k= 1 
coeff(term3, x, 1 )= 
               75*u^2*v^7-25*u^3*v^6-240*u^4*v^5+240*u^5*v^4+376*u^6*v^3
                         +120*u^7*v^2 
coeff(lb, t, 1 )= 299*u^5*v^5 
wdiff= 
      q*w
       *(75*q^4*w^14-25*q^6*w^12-240*q^8*w^10+240*q^10*w^8+376*q^12*w^6
                    +120*q^14*w^4)
       -299*q^10*w^10 
polynomial: 120*s^15+376*s^13+240*s^11-299*s^10-240*s^9-25*s^7+75*s^5 
factors= s^5*(120*s^10+376*s^8+240*s^6-299*s^5-240*s^4-25*s^2+75) 
number of roots >0: 0 
poly(2)= 7074016 
=== 
k= 2 
coeff(term3, x, 2 )= 
               75*u*v^7+165*u^2*v^6-675*u^3*v^5+1416*u^5*v^3+888*u^6*v^2
                       +120*u^7*v 
coeff(lb, t, 2 )= 834*u^5*v^5 
wdiff= 
      q^2*w^2
         *(75*q^2*w^14+165*q^4*w^12-675*q^6*w^10+1416*q^10*w^6+888*q^12*w^4
                      +120*q^14*w^2)
       -834*q^10*w^10 
polynomial: 120*s^16+888*s^14+1416*s^12-834*s^10-675*s^8+165*s^6+75*s^4 
factors= 3*s^4*(40*s^12+296*s^10+472*s^8-278*s^6-225*s^4+55*s^2+25) 
number of roots >0: 0 
poly(2)= 27198192 
=== 
k= 3 
coeff(term3, x, 3 )= 
               65*v^7+181*u*v^6-585*u^2*v^5-1286*u^3*v^4+2079*u^4*v^3
                     +2808*u^5*v^2+768*u^6*v+65*u^7 
coeff(lb, t, 3 )= 867*u^5*v^5 
wdiff= 
      q^3*w^3
         *(65*w^14+181*q^2*w^12-585*q^4*w^10-1286*q^6*w^8+2079*q^8*w^6
                  +2808*q^10*w^4+768*q^12*w^2+65*q^14)
       -867*q^10*w^10 
polynomial: 
           65*s^17+768*s^15+2808*s^13+2079*s^11-867*s^10-1286*s^9-585*s^7
                  +181*s^5+65*s^3 
factors= 
        s^3*(65*s^14+768*s^12+2808*s^10+2079*s^8-867*s^7-1286*s^6-585*s^4
                    +181*s^2+65) 
number of roots >0: 0 
poly(2)= 59331624 
=== 
k= 4 
coeff(term3, x, 4 )= 
               351*v^6-489*u*v^5-2058*u^2*v^4+546*u^3*v^3+4437*u^4*v^2
                      +2088*u^5*v+351*u^6 
coeff(lb, t, 4 )= -152*u^5*v^5 
wdiff= 
      q^4*w^4
         *(351*w^12-489*q^2*w^10-2058*q^4*w^8+546*q^6*w^6+4437*q^8*w^4
                   +2088*q^10*w^2+351*q^12)
       +152*q^10*w^10 
polynomial: 351*s^16+2088*s^14+4437*s^12+698*s^10-2058*s^8-489*s^6+351*s^4 
factors= s^4*(351*s^12+2088*s^10+4437*s^8+698*s^6-2058*s^4-489*s^2+351) 
number of roots >0: 0 
poly(2)= 75549104 
=== 
k= 5 
coeff(term3, x, 5 )= 
               741*v^5-2178*u*v^4-1476*u^2*v^3+3504*u^3*v^2+2997*u^4*v+741*u^5

coeff(lb, t, 5 )= -632*u^5*v^5 
wdiff= 
      q^5*w^5
         *(741*w^10-2178*q^2*w^8-1476*q^4*w^6+3504*q^6*w^4+2997*q^8*w^2
                   +741*q^10)
       +632*q^10*w^10 
polynomial: 741*s^15+2997*s^13+3504*s^11+632*s^10-1476*s^9-2178*s^7+741*s^5 
factors= s^5*(741*s^10+2997*s^8+3504*s^6+632*s^5-1476*s^4-2178*s^2+741) 
number of roots >0: 0 
poly(2)= 55645088 
=== 
k= 6 
coeff(term3, x, 6 )= 754*v^4-2698*u*v^3+1170*u^2*v^2+2412*u^3*v+754*u^4 
coeff(lb, t, 6 )= 2*u^5*v^5 
wdiff= 
      q^6*w^6*(754*w^8-2698*q^2*w^6+1170*q^4*w^4+2412*q^6*w^2+754*q^8)
       -2*q^10*w^10 
polynomial: 754*s^14+2412*s^12+1168*s^10-2698*s^8+754*s^6 
factors= 2*s^6*(377*s^8+1206*s^6+584*s^4-1349*s^2+377) 
number of roots >0: 0 
poly(2)= 22786688 
=== 
k= 7 
coeff(term3, x, 7 )= 390*v^3-1056*u*v^2+1134*u^2*v+390*u^3 
coeff(lb, t, 7 )= 531*u^5*v^5 
wdiff= q^7*w^7*(390*w^6-1056*q^2*w^4+1134*q^4*w^2+390*q^6)-531*q^10*w^10 
polynomial: 390*s^13+1134*s^11-531*s^10-1056*s^9+390*s^7 
factors= 3*s^7*(130*s^6+378*s^4-177*s^3-352*s^2+130) 
number of roots >0: 0 
poly(2)= 4482816 
=== 
k= 8 
coeff(term3, x, 8 )= 156*v^2-156*u*v+156*u^2 
coeff(lb, t, 8 )= 156*u^5*v^5 
wdiff= q^8*w^8*(156*w^4-156*q^2*w^2+156*q^4)-156*q^10*w^10 
polynomial: 156*s^12-312*s^10+156*s^8 
factors= 156*(s-1)^2*s^8*(s+1)^2 
number of roots >0: 2 
poly(2)= 359424 
(%o23) done
(%i24) "finally we proof that the lowerbbound polynomial has no positive root and that 

it is greater than 0 for t=1. Therefor it is greater or equal than 0 for all admissible values"
(%o24) "finally we proof that the lowerbbound polynomial has no positive root and that 

it is greater than 0 for t=1. Therefor it is greater or equal than 0 for all admissible values"
(%i25) poly:ratcoef(lowerbound,u^5*v^5)
(%o25) 156*t^8+531*t^7+2*t^6-632*t^5-152*t^4+867*t^3+834*t^2+299*t+40
(%i26) ev(poly,t = 1)
(%o26) 1945
(%i27) nroots(poly,0,inf)
(%o27) 0
(%i28) 

Here we list the coefficient functions so we can compare them to @MichaelRozenbergs function to see they are the same.

$$\begin{array}{r} \tag{1} \left(25\,u^3\,v^7-40\,u^4\,v^6+40\,u^6\,v^4+40\,u^7\,v^3\right)\,x^0 \\ \left(75\,u^2\,v^7-25\,u^3\,v^6-240\,u^4\,v^5+240\,u^5\,v^4+376\,u^ 6\,v^3+120\,u^7\,v^2\right)\,x^1 \\ \left(75\,u\,v^7+165\,u^2\,v^6-675\,u^3\,v^5+1416\,u^5\,v^3+888\,u^ 6\,v^2+120\,u^7\,v\right)\,x^2 \\ \left(65\,v^7+181\,u\,v^6-585\,u^2\,v^5-1286\,u^3\,v^4+2079\,u^4\,v ^3+2808\,u^5\,v^2+768\,u^6\,v+65\,u^7\right)\,x^3 \\ \left(351\,v^6-489\,u\,v^5-2058\,u^2\,v^4+546\,u^3\,v^3+4437\,u^4\, v^2+2088\,u^5\,v+351\,u^6\right)\,x^4 \\ \left(741\,v^5-2178\,u\,v^4-1476\,u^2\,v^3+3504\,u^3\,v^2+2997\,u^4 \,v+741\,u^5\right)\,x^5 \\ \left(754\,v^4-2698\,u\,v^3+1170\,u^2\,v^2+2412\,u^3\,v+754\,u^4 \right)\,x^6 \\ \left(390\,v^3-1056\,u\,v^2+1134\,u^2\,v+390\,u^3\right)\,x^7 \\ \left(156\,v^2-156\,u\,v+156\,u^2\right)\,x^8 \end{array}$$

To proof that this function is larger than $$\left(156\,t^8+531\,t^7+2\,t^6-632\,t^5-152\,t^4+867\,t^3+834\,t^2+ 299\,t+40\right)\,u^5\,v^5 \tag{2}$$ Rozenbergs's lower bound when we substitute $x$ by $t\sqrt(uv)$ we show that each coefficient of the polynomial $(1)$ is larger than the corresponding coefficient of the lower bound polynomial $(2)$. Then we show that the polynomial $(2)$ is larger than $0$ for all nonnegative $u$, $v$ and $t$. Details can be found in the Maxima script.

Instead of the Maxima nroots function, which is based on Sturm sequences, one could solve the equations by some numeric functions to see if there are zeros greater than zeros, e.g. calculating the roots of poly for k=7 gives the following:

(%i29) allroots(390*s^13+1134*s^11-531*s^10-1056*s^9+390*s^7 ,s);
(%o29) [s = 0.0,s = 0.0,s = 0.0,s = 0.0,s = 0.0,s = 0.0,s = 0.0,
        s = 0.007444635413686057*%i+0.7516683014652126,
        s = 0.7516683014652126-0.007444635413686057*%i,
        s = 0.3202741285237583*%i-0.6047586795035632,
        s = (-0.3202741285237583*%i)-0.6047586795035632,
        s = 1.93839678615644*%i-0.1469096219616494,
        s = (-1.93839678615644*%i)-0.1469096219616494]

So we can also conclude are no real roots greater than 0. But this method is not really acceptable if one does not analyze the impact of the rounding errors. And this can be very complicated. The nroots function works with integers (for integer polynomials) and so there are no rounding errors.

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6
$\begingroup$

I write a start for a full answer (this is an idea that @Starfall first proposed in comment). If someone wants to use it to end the proof, she/he is welcome!

Let $$f(x,y,z):=\frac{x^4}{ax^3+by^3}+\frac{y^4}{ay^3+bz^3}+\frac{z^4}{az^3+bx^3}.$$ Since $f$ is homogeneous of degree 1, it is sufficient to consider $x,y,z$ on the plane $P:=\{x+y+z=1\}$. Let $$g(x,y,z):=x+y+z-1$$ be the constraint function. We compute : $$\mathrm{d}f(x,y,z)=\left(\frac{ax^6+4bx^3y^3}{(ax^3+by^3)^2}-\frac{3bx^2z^4}{(az^3+bx^3)^2}\right)\mathrm{d}x+\left(\frac{ay^6+4by^3z^3}{(ay^3+bz^3)^2}-\frac{3bx^4y^2}{(ax^3+by^3)^2}\right)\mathrm{d}y$$ $$+\left(\frac{az^6+4bx^3z^3}{(az^3+bx^3)^2}-\frac{3by^4z^2}{(ay^3+bz^3)^2}\right)\mathrm{d}z,$$ $$\mathrm{d}g(x,y,z)=\mathrm{d}x+\mathrm{d}y+\mathrm{d}z.$$ Define the $2\times 3$ matrix $$M:=\begin{pmatrix} \frac{\partial f}{\partial x}(x,y,z) & \frac{\partial f}{\partial y}(x,y,z) & \frac{\partial f}{\partial z}(x,y,z)\\ \frac{\partial g}{\partial x}(x,y,z) & \frac{\partial g}{\partial y}(x,y,z) & \frac{\partial g}{\partial z}(x,y,z) \end{pmatrix}.$$ By Lagrange multipliers theorem, all the 3 sub-determinants of $M$ must vanish at a local minimum $(x,y,z)$ of $f$ on $P$.

Setting $$A:=ax^3+by^3,\quad B:=az^3+bx^3,\quad ay^3+bz^3,$$ cancelling the 3 sub-determinants of $M$ yields : \begin{align} \begin{cases} B^2C^2(ax^6+4bx^3y^3+3bx^4y^2)-3A^2C^2bx^2z^4-A^2B^2(ay^6+4by^3z^3)&=0\\ B^2C^2(ax^6+4bx^3y^3)-A^2C^2(3bx^2z^4+az^6+4bx^3z^3)+3A^2B^2by^4z^2&=0\\ A^2B^2(ay^6+4by^3z^3+3by^4z^2)-3B^2C^2bx^4y^2-A^2C^2(az^6+4bx^3z^3)&=0\\ x+y+z=1,\ x,y,z>0 \end{cases}. \end{align} Labelling the lines $(1)$, $(2)$, $(3)$ and $(4)$, we can see that $(1)-(2)=-(3)$, so that we can forget one of the three first lines.

Here we need to do some (boring) algebra, using the constraints of the fourth line above and maybe some tricks like writing $ax^3=A-by^3$ and $bx^4=(1-y-z)(B-az^3)$. But I am too busy now to try this, and I don't know if I would try later...

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2
  • $\begingroup$ Thank you very much for using Lagrange Multiplier ! One up vote from me :) $\endgroup$
    – HN_NH
    May 24, 2016 at 6:06
  • 1
    $\begingroup$ @HN_NH Thank you! Unfortunately, using the Lagrange multipliers theorem just yields a horrible system of algebraic equations. You should be able to solve it with a quite powerful program. $\endgroup$
    – Nicolas
    May 24, 2016 at 9:07
6
+500
$\begingroup$

Another way.

By C-S $$\sum_{cyc}\frac{x^4}{8x^3+5y^3}=\sum_{cyc}\frac{x^4(3x-y+2z)^2}{(8x^3+5y^3)(3x-y+2z)^2}\geq\frac{\left(\sum\limits_{cyc}(3x^3-x^2y+2x^2z)\right)^2}{\sum\limits_{cyc}(8x^3+5y^3)(3x-y+2z)^2}.$$ Thus, it's enough to prove that: $$13\left(\sum\limits_{cyc}(3x^3-x^2y+2x^2z)\right)^2\geq(x+y+z)\sum\limits_{cyc}(8x^3+5y^3)(3x-y+2z)^2.$$ Since the last inequality is cyclic, we can assume that $x=\min\{x,y,z\}$.

  1. Let $x\leq z\leq y$, $z=x+u$ and $y=x+u+v$.

Thus, $u$ and $v$ are non-negatives and we need to prove that: $$166(u^2+uv+v^2)x^4+(555u^3+1791u^2v+1454uv^2+109v^3)x^3+$$ $$+(861u^4+3639u^3v+4284u^2v^2+1506uv^3+192v^4)x^2+$$ $$+(555u^5+2474u^4v+3833u^3v^2+2317u^2v^3+153uv^4+166v^5)x+$$ $$+123u^6+547u^5v+1046u^4v^2+843u^3v^3+374u^2v^4+153uv^5+40v^6\geq0,$$ which is obvious;

  1. Let $x\leq y\leq z,$ $y=x+u$ and $z=x+u+v$.

Thus, we need to prove that: $$166(u^2+uv+v^2)x^4+(555u^3-126u^2v-463uv^2+109v^3)x^3+$$ $$+(861u^4-195u^3v-1467u^2v^2-411uv^3+192v^4)x^2+$$ $$+(555u^5+301u^4v-513u^3v^2-112u^2v^3+479uv^4+166v^5)x+$$ $$+123u^6+191u^5v+156u^4v^2+331u^3v^3+496u^2v^4+253uv^5+40v^6\geq0.$$ Easy to show that: $$166(u^2+uv+v^2)\geq498uv,$$ $$555u^3-126u^2v-463uv^2+109v^3\geq-249\sqrt{u^3v^3},$$ $$861u^4-195u^3v-1467u^2v^2-411uv^3+192v^4\geq-1494u^2v^2,$$ $$555u^5+301u^4v-513u^3v^2-112u^2v^3+479uv^4+166v^5\geq747\sqrt{u^5v^5}$$ and $$123u^6+191u^5v+156u^4v^2+331u^3v^3+496u^2v^4+253uv^5+40v^6\geq1494u^3v^3.$$ Thus, after substitution $x=t\sqrt{uv}$ it's enough to prove that $$498t^4-249t^3-1494t^2+747t+1494\geq0,$$ which is true because $$498t^4-249t^3-1494t^2+747t+1494=$$ $$=249(t+1)(2t^3-3t^2-3t+6)=249(t+1)(t^3+2-3t+t^3+4-3t^2)\geq$$ $$\geq249(t+1)\left(3\sqrt[3]{t^3\cdot1^2}-3t+3\sqrt[3]{\left(\frac{t^3}{2}\right)^2\cdot4}-3t^2\right)=0.$$ Done!

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6
  • $\begingroup$ +1 ; about time you answered it by Holder with polynomial integers :> $\endgroup$
    – user552223
    Jun 7, 2020 at 10:31
  • $\begingroup$ You did it by your algorithm, didn't you ? $\endgroup$
    – user552223
    Jun 7, 2020 at 10:33
  • 1
    $\begingroup$ @Giang Nguyễn Đặng Thanh I just played with coefficients and it turned out. $\endgroup$ Jun 7, 2020 at 10:48
  • 1
    $\begingroup$ @Giang Nguyễn Đặng Thanh I tried, but without success. I posted a proof of this inequality here: math.stackexchange.com/questions/3589716 $\endgroup$ Jun 7, 2020 at 12:38
  • 2
    $\begingroup$ (+1) Nice C-S, now the form is better. $\endgroup$
    – River Li
    Jun 7, 2020 at 14:19
3
$\begingroup$

Not sure, if I missed out anything here. Take a look.

For non negative, $X,Y,Z$, We can perhaps use Titu's inequality (a mix of Holder and CS), sometimes called Titu's screw lemma (https://en.wikipedia.org/wiki/Nesbitt%27s_inequality). \begin{equation} \sum_{k=1}^{n}{\frac{x_{k}^{2}}{a_{k}}} \ge \frac{\left(\sum_{k=1}^{n}{x_{k}}\right)^{2}}{\sum_{k=1}^{n}{a_{k}}} \end{equation}

With $n\to3$ terms, $x_{1}\to X^{2},x_{2} \to Y^{2}, x_{3} \to Z^{2}$ and $a_{1} \to A, a_{2}\to B, a_{3} \to C$, we will have

\begin{eqnarray*} \frac{\left(X^2\right)^{2}}{A}+\frac{\left(Y^2\right)^{2}}{B}+\frac{\left(Z^2\right)^{2}}{C} &\ge& \frac{\left(X^{2}+Y^{2}+Z^{2}\right)^{2}}{A+B+C} \\ \end{eqnarray*}

With \begin{eqnarray*} A &=& \alpha X^{3} +\beta Y^{3} \\ B &=& \alpha Y^{3} +\beta Z^{3} \\ C &=& \alpha Z^{3} +\beta X^{3} \end{eqnarray*}

where, \begin{eqnarray*} A+B+C &=& (\alpha+\beta) \left(X^{3} + Y^3+Z^3 \right) \end{eqnarray*}

\begin{eqnarray} \frac{X^4}{A}+\frac{Y^4}{B}+\frac{Z^4}{C} &=&\frac{\left(X^2\right)^{2}}{A}+\frac{\left(Y^2\right)^{2}}{B}+\frac{\left(Z^2\right)^{2}}{C}\\ &\ge& \frac{\left(X^{2}+Y^{2}+Z^{2}\right)^{2}}{A+B+C} \\ &=& \frac{\left(X^{2}+Y^{2}+Z^{2}\right)^{2}}{(\alpha+\beta) \left(X^{3} + Y^3+Z^3 \right)} \\ &\overset{(p)}{\ge}& \frac{\left(X^{3}+Y^{3}+Z^{3}\right)\left(X+Y+Z\right)}{(\alpha+\beta) \left(X^{3} + Y^3+Z^3 \right)} \\ &=& \frac{\left(X+Y+Z\right)}{(\alpha+\beta)} \end{eqnarray}

QED.

Here $(p)$ is from the fact that,

\begin{eqnarray*} (X^2+Y^2+Z^2)^{2} -\left(X^{3}+Y^{3}+Z^{3} \right) (X+Y+Z) &=& XY(X-Y)^{2}+YZ(Y-Z)^{2}+ZX(Z-X)^{2} \\ &\ge& 0 \end{eqnarray*}

Here $\alpha=8$ and $\beta=5$.

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6
  • $\begingroup$ By the way, I just used the name Titu (as it is known in the US IMO circle). I also hear names such as T2 Lemma, Engel's form, or Sedrakyan's inequality. It is indeed a special case of Cauchy-Schwarz. $\endgroup$ Aug 22, 2018 at 16:22
  • $\begingroup$ Yes, it is also known by the name Bergstrom’s inequality. There is also a slightly tighter generalization as well. $\endgroup$ Aug 22, 2018 at 16:35
  • 1
    $\begingroup$ I think the identity for (p) is wrong: $(1^2+1^2+2^2)^2-(1^3+1^3+2^3)(1+1+2)=36-40=-4$. When you expand, you should get $2X^2Y^2-X^3Y-XY^3=-XY(X-Y)^2.$ $\endgroup$
    – Jose Brox
    Aug 22, 2018 at 16:37
  • $\begingroup$ Thanks @Jose Brox. Yes, then there is a hole in my argument. We still have it open then:-) $\endgroup$ Aug 22, 2018 at 17:04
  • $\begingroup$ @ Michael Rozenberg, I didn't mean so:-) I only meant to say, using this method:-). I still have feel that, there is a way to prove using T2 (Mostly, we may have to include the additional term involving max and then show that it stays positive). Will take a look over the weekend. $\endgroup$ Aug 22, 2018 at 17:20
3
+50
$\begingroup$

For checking purposes.

Making $y = \lambda, \ z = \mu x$ and substituting into

$$ f(x,y,z) = \frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} - \frac{x+y+z}{13} $$

giving

$$ g(x,\lambda,\mu) =x\left( \frac{1}{5 \lambda ^3+8}+\frac{\lambda ^4}{8 \lambda ^3+5 \mu ^3}+\frac{\mu ^4}{8 \mu ^3+5}-\frac{1}{13} (\lambda +\mu +1)\right) $$

and discarding $x > 0$ we get

$$ \mathcal{G}(\lambda,\mu) = \frac{1}{5 \lambda ^3+8}+\frac{\lambda ^4}{8 \lambda ^3+5 \mu ^3}+\frac{\mu ^4}{8 \mu ^3+5}-\frac{1}{13} (\lambda +\mu +1) $$

Now solving the stationary conditions

$$ \nabla\mathcal{G}(\lambda,\mu) = 0 $$

we have the feasible stationary points with qualification.

$$ \left[ \begin{array}{cccl} \lambda & \mu & \mathcal{G}(\lambda,\mu) & \mbox{kind} \\ 1. & 1. & 0. & \mbox{min} \\ 0.485435 & 0.715221 & 0.000622453 & \mbox{min}\\ 0.646265 & 0.811309 & 0.000758688 & \mbox{saddle} \\ 1.37554 & 0.688678 & 0.000863479 & \mbox{min} \\ 1.25 & 0.77611 & 0.000941355 & \mbox{saddle} \\ 1.38778 & 1.85522 & 0.00123052 & \mbox{min} \\ 1.34211 & 1.74761 & 0.00123288 & \mbox{saddle} \\ \end{array} \right] $$

so the best solution is at $x = y = z = 1$

Attached the level contours for $\mathcal{G}(\lambda,\mu)$ with the stationary points in red.

enter image description here

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    $\begingroup$ It's not a solution, of course. How you solved this system? How you got a minimal point? If you used computer, so it's just nothing because WA says that the starting inequality is true, but it also is not a proof. @quid♦ Explain us please, why did you accept this answer? $\endgroup$ Jan 5, 2020 at 3:21
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Let's reform this inequality in a way such that we can comprehend it better. Define $a=\dfrac{y}{x}$ and $b=\dfrac{z}{y}$, therefore $\dfrac{x}{z}={1\over ab}$. We can suume without lose of generality that $a,b\le1$ We need to prove that $$\dfrac{x}{8+5\left(\dfrac{y}{x}\right)^3}+\dfrac{y}{8+5\left(\dfrac{z}{y}\right)^3}+\dfrac{z}{8+5\left(\dfrac{x}{z}\right)^3}\ge\dfrac{x+y+z}{13}$$by dividing the two sides of the inequality by $x$ and substituting $a,b,c$ we have that$$\dfrac{1}{8+5\left(\dfrac{y}{x}\right)^3}+\dfrac{\dfrac{y}{x}}{8+5\left(\dfrac{z}{y}\right)^3}+\dfrac{\dfrac{z}{y}}{8+5\left(\dfrac{x}{z}\right)^3}\ge\dfrac{1+\dfrac{y}{x}+\dfrac{z}{x}}{13}$$and $$\dfrac{1}{8+5a^3}+\dfrac{a}{8+5b^3}+\dfrac{a^4b^4}{5+8a^3b^3}\ge \dfrac{1}{13}+\dfrac{a}{13}+\dfrac{ab}{13}$$which is equivalent to $$\left(\dfrac{1}{8+5a^3}-\dfrac{1}{13}\right)+\left(\dfrac{a}{8+5b^3}-\dfrac{a}{13}\right)+\left(\dfrac{a^4b^4}{5+8a^3b^3}-\dfrac{ab}{13}\right)\ge 0$$by simplifying each of the components and multiplying both sides in $\dfrac{13}{5}$ we obtain$$\dfrac{1-a^3}{8+5a^3}+\dfrac{a(1-b^3)}{8+5b^3}+\dfrac{a^4b^4-ab}{5+8a^3b^3}\ge0$$below is a depiction of $f(a,b)=\dfrac{1-a^3}{8+5a^3}+\dfrac{a(1-b^3)}{8+5b^3}+\dfrac{a^4b^4-ab}{5+8a^3b^3}$ for $0\le a,b\le 1$

enter image description here

which proves the inequality graphically (I believe that Lagrange multipliers or any other method based on 1st order derivations may help but i hadn't much time to think on it hope you find an analytic way) but neither such a time i spent on the problem nor a computer is given us in the exam :) also i appreciate if any one updates his/her comment with such an analytical method. I'm really curious about that.....

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I have finally found a solution . In fact we start to study the 2 variables version of this inequality we have :

$$\frac{a^4}{8a^3+5b^3}+\frac{b^4}{8b^3+5a^3}\geq \frac{a+b}{13}$$

Proof:

We have with $x=\frac{a}{b}$ : $$\frac{x^4}{8x^3+5}+\frac{1}{8+5x^3}\geq \frac{1+x}{13}$$ Or $$\frac{5}{13}(x - 1)^2 (x + 1) (x^2 + x + 1) (5 x^2 - 8 x + 5)\geq 0$$

So we have (if we permute the variables $a,b,c$ and add the three inequalities ) :

$$\sum_{cyc}\frac{a^4}{8a^3+5b^3}+\sum_{cyc}\frac{a^4}{8a^3+5c^3}\geq \frac{a+b+c}{6.5}$$

If we have $\sum_{cyc}\frac{a^4}{8a^3+5b^3}\geq\sum_{cyc}\frac{a^4}{8a^3+5c^3}$

We have : $$\sum_{cyc}\frac{a^4}{8a^3+5b^3}\geq \frac{a+b+c}{13}$$ But also $$\frac{(a-\epsilon)^4}{8(a-\epsilon)^3+5b^3}+\frac{(b)^4}{8(b)^3+5(c+\epsilon)^3}+\frac{(c+\epsilon)^4}{8(c+\epsilon)^3+5(a-\epsilon)^3}\geq \frac{a+b+c}{13}$$ If we put $a\geq c $ and $\epsilon=a-c$

We finally obtain : $$\sum_{cyc}\frac{a^4}{8a^3+5c^3}\geq \frac{a+b+c}{13}$$

If we have $\sum_{cyc}\frac{a^4}{8a^3+5b^3}\leq\sum_{cyc}\frac{a^4}{8a^3+5c^3}$

The proof is the same as above .

So all the cases are present so it's proved !

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$\color{green}{\textbf{Light version (24.01.20).}}$

$\color{brown}{\textbf{Inequalities for cubic root.}}$

Searching of the polynomials in the forms of \begin{cases} P_4(s)=s(1+as^3) - (1+a)s^3 = as^4 -(a+1)s^3+s\\[4pt] P_7(s)=(5+8s^3)(1-b+bs^3) - s(13-c+cs^3)\\ \qquad = 8bs^6-cs^4+(8-3b)s^3+(c-13)s+5-5b \end{cases} under the conditions $$P'_4(1)=P'_7(1)=P''_7(1) = 0,$$ allows to obtain the coefficients $a,b,c:$ $$ \begin{cases} 4a-3(1+a)+1=0\\ 48b-4c+3(8-3b)+c-13=0\\ 240b-12c+6(8-3b)=0 \end{cases} \Rightarrow \begin{cases} a = 2\\ 39b-3c = -11\\ 222b-12c = -48, \end{cases} $$ $$a=2,\quad b=-\dfrac2{33},\quad c=\dfrac{95}{33},$$

then \begin{cases} P_4(s) = s(1+2s^3) - 3s^3 = s(1-s)^2(2s+1)\\ 33P_7(b)=(35-2s^3)(5+8s^3) - s(334+95s^3) = (1-s)^3(16s^3+48s^2+191s+175). \end{cases} If $s\in[0,1]\ $ then $P_4(s)\ge0,\ P_7(s)\ge0.$

Applying of the substitution $s=\sqrt[\large 3]{1-t\large\mathstrut}\ $ leads to the inequalities

$$\dfrac{(13-8t)(33+2t)}{429-95t} \ge \sqrt[\large3]{1-t\mathstrut} \ge \dfrac{3(1-t)}{3-2t}\quad\text{if} \quad t\in[0,1]\tag1$$

(see also Wolfram Alpha plot).

Inequalities for cubic root

On the other hand, the function $$S(t)=\sqrt[\Large3]{\dfrac{5t\mathstrut}{13-8t}},\quad t\in[0,1]$$

has the inverse one in the form of $$T(s)=\dfrac{13s^3}{8s^3+5},\quad s\in[0,1].$$

If $s=S(t),$ then \begin{align} &\dfrac{15+11t-11t^2}{3(13-8t)}-S(t) = \dfrac{15+11T(s)-11T^2(s)}{3(13-8T(s))}-s\\[8pt] & = \dfrac{49s^6-312s^4+383s^3-195s+75}{312s^2+195} = \dfrac{(s+1)^2(7s+5)(7s^3+9s^2-30s+15)}{39(8s^2+5)},\\[8pt] &7s^3+9s^2-30s+15 = 7(1-s)(1-s^2)+8(1-s)(2-s)+s, \end{align}

$$S(t) = \sqrt[\Large3]{\dfrac{5t\mathstrut}{13-8t}} \le \dfrac{15+11t-11t^2}{3(13-8t)},\quad t\in[0,1].\tag2$$

(see also Wolfram Alpha plot).

Cubic root approximation, plot2

$\color{brown}{\textbf{Primary transformations.}}$

The given inequality WLOG can be presented in the forms of $$x\ge y,\quad x\ge z,\quad \dfrac{x^4}{8x^3+5y^3}+\dfrac{y^4}{8y^3+5z^3}+\dfrac{z^4}{8z^3+5x^3} \ge \dfrac1{13}(x+y+z),\tag3$$

or $$\dfrac{13x^4}{8x^3+5y^3}-x + \dfrac{13y^4}{8y^3+5z^3}-y + \dfrac{13z^4}{8z^3+5x^3}-z \ge 0,$$

$$\dfrac{x^3-y^3}{8x^3+5y^3} + \dfrac yx\,\dfrac{y^3-z^3}{8y^3+5z^3} - \dfrac zx\,\dfrac{x^3-z^3}{5x^3+8z^3} \ge 0.\tag4$$

$\color{brown}{\mathbf{Case\ \ z < y \le x.}}$

Taking in account $(1),$ inequality $(4)$ in the notation $$\dfrac{z^3}{x^3} = 1-u,\quad \dfrac{y^3}{x^3} = 1-uv,\quad (u,v)\in[0,1]^2, \tag5$$ \begin{align} &\dfrac{x^3-y^3}{8x^3+5y^3} = \dfrac{uv}{8+5(1-uv)},\quad \dfrac yx = \sqrt[\large3]{1-uv\mathstrut},\\[8pt] &\dfrac{y^3-z^3}{8y^3+5z^3} = \dfrac{u-uv}{8(1-uv)+5(1-u)},\\[8pt] &\dfrac{x^3-z^3}{5x^3+8z^3} = \dfrac{u}{5+8(1-u)},\quad \dfrac zx = \sqrt[\large3]{1-u\mathstrut}, \end{align}

takes the form of $f_1(u,v) \ge 0,$ where \begin{align} &f_1(u,v) = u\left(\dfrac{v}{13-5uv} + \dfrac{3(1-uv)}{3-2uv}\,\dfrac{1-v}{13-5u-8uv} - \dfrac{33+2u}{429-95u}\right)\\[8pt] & = \dfrac{u^2(A(u)+vB(u)+v^2C(u)+v^3D(u))}{(3-2uv)(13-5u-8uv)(13-5uv)(429-95u)}, \end{align} \begin{align} & A(u) = 1716+390u,\\ & B(u) = -1716+1480u-410u^2,\\ & C(u) = 1716-4769u-1641u^2+100u^3,\\ & D(u) = 429u + 2545u^2+160u^3,\\ & A(u)+vB(u)+v^2C(u)+v^3D(u) = (1-v)(1-v^2)A(u)+v(1-v)^2(A(u)+B(u))\\ & +v^2(1-v)(3A(u)+2B(u)+C(u))+v^3(A(u)+B(u)+C(u)+D(u))\\ & = (1-v)(1-v^2)(1716+390u)+v(1-v)^2(1870u-410u^2)\\ & +v^2(1-v)(3432-639u-2461u^2+100u^3)+26v^3(1-u)(66-29u-10u^2) \ge 0 \end{align} (see also Wolfram Alpha checking and matrix calculations).

Input $\!\mathstrut^{\phantom{\dfrac\mathstrut\mathstrut}^{\LARGE=}}$ Result

Therefore, $f_1(u,v)\ge0.$

The case is proved.

$\color{brown}{\mathbf{Case\ \ y \le z \le x.\ Additional\ transformations.}}$

Using the notation $$\dfrac{5(x^3-z^3)}{5x^3+8z^3} = 1-u,\quad \dfrac{5(z^3-y^3)}{5z^3+8y^3} = 1-v,\quad (u,v)\in[0,1]^2, \tag6$$

one can get $$\dfrac{z^3}{x^3} = \dfrac{5u}{13-8u},\quad\dfrac{y^3}{z^3} = \dfrac{5v}{13-8v},\quad \dfrac{y^3}{x^3} = \dfrac{25uv}{(13-8u)(13-8v)},$$ $$\dfrac{x^3-y^3}{8x^3+5y^3} = \dfrac{(13-8u)(13-8v)-25uv}{8(13-8u)(13-8v)+125uv} = \dfrac{13-8(u+v)+3uv}{104-64(u+v)+49uv}.\tag7$$

$\color{brown}{\mathbf{Case\ \ y \le z \le x,\ u+v \ge \dfrac{13}8.}}$

Taking in account $(2),$ the inequality $(4)$ takes the stronger form of $f_2(u,v)\ge0,$ where \begin{align} &f_2(u,v) = 5\dfrac{13-8(u+v)+3uv}{104-64(u+v)+49uv}- (1-v)S(u)S(v) - (1-u)S(u)\\[8pt] & \ge 5\dfrac{13-8(u+v)+3uv}{104-64(u+v)+49uv} - \dfrac{15+11u-11u^2}{3(13-8u)} \left((1-v)\dfrac{15+11v-11v^2}{3(13-8v)}+1-u\right)\\ & = \dfrac{g_2(u,v)}{9(104-64(u+v)+49uv)(13-8u)(13-8v)},\\[8pt] &g_2(u,v)=5(13-8(u+v)+3uv)(39-24u)(39-24v)\\[4pt] &-((1-v)(15+11(1-v)v)+(39-24v)(1-u))\\[4pt] &\times(15+11(1-u)u)(104-64(u+v)+49uv). \end{align}

Let $p=1-u,\ \ q=1-v,$ then $p+q \in \left[0,\dfrac58\right],$

\begin{align} &g^\,_2(p,q) = 5(5(p+q)+3pq)(15+24p)(15+24q)\\[4pt] &-(q(15+11(1-q)q)+(15+24q)p)(15+11(1-p)p)(25+15(p+q)+49pq)\\[4pt] &= 1500p^2+1500pq+1500q^2\\[4pt] &+1650p^3-4050p^2q-4600pq^2+1650q^3\\[4pt] &+2475p^4-495p^3q-17360p^2q^2-4400pq^3+2475q^4\\[4pt] &+12045p^4q+924p^3q^2-5324p^2q^3+9900pq^4\\[4pt] &+12936p^4q^2+4114p^3q^3+4114p^2q^4-5929p^3q^4\\[4pt] \end{align} (see also Wolfram Alpha checking).

Since $$pq \le \dfrac14(p+q)^2,\quad p^3-p^2q-pq^2+q^3 = (p-q)(p^2-q^2) \ge 0,$$

then \begin{align} &g^\,_2(p,q) \ge 375(4(p+q)^2-4pq)\\[4pt] & + 1650(p-q)(p^2-q^2) - 3000pq(p+q)\\[4pt] &+2475(p^2-q^2)^2 -pq(495p^2+12410pq+4400q^2)\\[4pt] &+9900pq(p-q)(p^2-q^2)\\[4pt] &+4114p^2q^3(p(1-p)+ q(1-q)+p^2+q^2-2pq)\\[4pt] &\ge 1125(p+q)^2-3000pq(p+q)-6208pq(p+q)^2 + 0 + 0\\[4pt] &\ge 1125(p+q)^2-750(p+q)^3-1552(p+q)^4\\[4pt] &\ge \left(1125 - 750\cdot\dfrac58-1552\cdot\dfrac{25}{64}\right)(p+q)^2 \ge 0. \end{align}

The case is proved.

$\color{brown}{\mathbf{Case\ \ y \le z \le x,\ u+v \le \dfrac{13}8.}}$

From $(7)$ should \begin{align} &\dfrac{(49\, \dfrac{x^3-y^3}{8x^3+5y^3}-3)}{100} = \dfrac{13-8(u+v)}{416-256(u+v)+49(2\sqrt{uv})^2} \ge \dfrac{13-8(u+v)}{416-256(u+v)+49(u+v)^2}. \end{align}

Since $$\dfrac1{49}\left(100\dfrac{13-8t}{416-256t+49t^2}+3\right) = \dfrac{(2-t)(26-3t)}{416-256t+49t^2}$$

and $$\dfrac{26-3t}{416-256t+49t^2} - \dfrac1{800}(50+21t+17t^2) = \dfrac{t(2-t)(833t^2-1657t+832)}{800(49t^2-256t+416)}$$ (see also Wolfram Alpha plot),

Ratio plot

then $$\dfrac{x^3-y^3}{8x^3+5y^3}\ge R(u+v),$$ where

$$R(t) = \dfrac1{800}(2-t)(50+21t+17t^2),\quad t\in[0,2].\tag8$$

Therefore, the inequality $(3)$ takes the stronger form of $f_3(u,v)\ge0,$ where \begin{align} &f_3(u,v) = 5R(u+v)- (1-v)S(u)S(v) - (1-u)S(u)\\[8pt] & \ge \dfrac{2-u-v}{160}(50+21(u+v)+17(u+v)^2)\\[8pt] & - \dfrac{15+11u-11u^2}{3(13-8u)} \left((1-v)\dfrac{15+11v-11v^2}{3(13-8v)}+1-u\right)\\ & = \dfrac{g_3(u,v)}{1440(13-8u)(13-8v)},\\[8pt] \end{align}

where \begin{align} &g^\,_3(u,v) = (50+21(u+v)+17(u+v)^2)(2-u-v)(39-24u)(39-24v)\\[4pt] &-160((1-v)(15+11(1-v)v)+(39-24v)(1-u))(15+11(1-u)u),\\[4pt] &g^\,_3(1-u,1-v) = (160-89(u+v)+17(u+v)^2)(u+v)(15+24u)(15+24v)\\[4pt] & - 160(15+11(1-u)u)((15+11(1-v)v)v+u(15+24v))\\[4pt] &= 11175u^2-1815u^3+6120u^4-8850uv-8325u^2v+15456u^3v+9792u^4v\\[4pt] &+11175v^2-11845uv^2-46448u^2v^2+29376u^3v^2\\[4pt] &-1815v^3-7424uv^3+10016u^2v^3+6120v^4+9792uv^4 \end{align} (see also Wolfram Alpha checking).

In the matrix form, $$ g^\,_3(1-u,1-v) = \mu(u,v,G_3) = \begin{pmatrix} 1 \\ v \\ v^2 \\ v^3 \\ v^4 \end{pmatrix}^T G_3 \begin{pmatrix} 1 \\ u \\ u^2 \\ u^3 \\ u^4 \end{pmatrix},\tag9 $$

where $$G_3 = \begin{pmatrix} 0 & 0 & 11175 & -1815 & 6120 \\ 0 & -8850 & -8325 & 15456 & 9792 \\ 11175 & -11845 & -46448 & 29376 & 0 \\ -1815 & -7424 & 10016 & 0 & 0 \\ 6120 & 9792 & 0 & 0 & 0 \end{pmatrix}.\tag{10} $$

At the same time:

  • $$ (u-v)^2(1-u-v)^2 = \begin{pmatrix} 1 \\ v \\ v^2 \\ v^3 \\ v^4 \end{pmatrix}^T \begin{pmatrix} 0 & 0 & 1 & -2 & 1 \\ 0 &-2 & 2 & 0 & 0 \\ 1 & 2 & -2 & 0 & 0 \\ -2 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ \end{pmatrix} \begin{pmatrix} 1 \\ u \\ u^2 \\ u^3 \\ u^4 \end{pmatrix}, $$

  • $$g_3(u,v) = 6120(u-v)^2(1-u-v)^2 + uv(9792(u-v)(u^2-v^2)+15456(u-v)^2)\\ + g^\,_{32}(u,v) = g^\,_{30}(u,v) + g^\,_{31}(u,v) + g^\,_{32}(u,v) = \mu(u,v,G_{30}+G_{31}+G_{32}),$$ where $$G_{30} = \begin{pmatrix} 0 & 0 & 6120 & -12240 & 6120 \\ 0 & -12240 & 12240 & 0 & 0 \\ 6120 & 12240 & -12240 & 0 & 0 \\ -12240 & 0 & 0 & 0 & 0 \\ 6120 & 0 & 0 & 0 & 0 \end{pmatrix}, $$ $$G_{31} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 15456 & 9792 \\ 0 & 0 & -30912 & -9792 & 0 \\ 0 & -15456 & -9792 & 0 & 0 \\ 0 & 9792 & 0 & 0 & 0 \end{pmatrix}, $$ $$G_{32} = \begin{pmatrix} 0 & 0 & 5055 & 10425 & 0 \\ 0 & 3390 & -3915 & 0 & 0 \\ 5055 & -24085 & -3296 & 39168 & 0 \\ 10425 & -22880 & 19808 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}, $$$$ g^\,_{30}(u,v) \ge 0,\quad g^\,_{31}(u,v) \ge 0. $$

Since

Input $\!\mathstrut^{\phantom{\dfrac\mathstrut\mathstrut}^{\LARGE=}}$ Result $\!\mathstrut^{\phantom{\dfrac\mathstrut\mathstrut}^{\LARGE,}}$

then, similarly to the first case, $$g^\,_{32}(u,v)= \begin{pmatrix} 1 \\ v \\ v^2 \\ v^3 \end{pmatrix}^T \begin{pmatrix} 0 & 0 & 5055 & 10425 \\ 0 & 3390 & -3915 & 0 \\ 5055 & -24085 & -3296 & 39168 \\ 10425 & -22880 & 19808 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ u \\ u^2 \\ u^3 \end{pmatrix}\\ =\begin{pmatrix} (1-v)(1-v^2) \\ v(1-v)^2 \\ v^2(1-v) \\ v^3 \end{pmatrix}^T \begin{pmatrix} 0 & 0 & 5055 & 10425 \\ 0 & 3390 & 1140 & 10425 \\ 5055 & -17305 & 40039 & 70443 \\ 15480 & -43575 & 17652 & 49593 \end{pmatrix} \begin{pmatrix} 1 \\ u \\ u^2 \\ u^3 \end{pmatrix}, $$ wherein

$$5055 - 17305u + 4039u^2 + 70443u^3 = 5055(1-2u)^2 + u(2915-16181u+70443) \ge 0,$$ $$15480 - 43575u + 17652u^2 + 49593u^3 = 15480(1-2u)^2 +3u(6115 -14756u + 16531u^2) \\ \ge0,$$ because the quadratic polynomials have negative discriminants (see also Wolfram Alpha plot).

Cubic polynomials plot

Thus, $g^\,_{32}(u,v)\ge 0$ and $g_3(u,v) \ge 0.$

PROVED.

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    $\begingroup$ Dear @Yuri Negometyanov I don't see a connection between your paper and starting inequality. I am sorry. $\endgroup$ Jun 22, 2017 at 2:40
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    $\begingroup$ @YuriNegometyanov You claim that $a$ achieves its maximum when $\lim \mathrm{LHS} = \lim \mathrm{RHS}$, right? I think that $\lim \mathrm{LHS} = \lim \mathrm{RHS}$ tells nothing about the maximum of $a$. $\endgroup$
    – River Li
    Jan 3, 2020 at 3:55
  • $\begingroup$ @RiverLi There are 5 groups of terms of $g_2$ (one group in one string), before "Since..." and 5 corresponding groups after it. All of groups contain 3-5 terms, and transformations can be easily checked. I think, we can return to $k=^{133}\!/_{81}.$ $\endgroup$ Jan 25, 2020 at 15:51
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    $\begingroup$ @YuriNegometyanov Please explain the 3rd group. $\endgroup$
    – River Li
    Jan 25, 2020 at 16:19
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    $\begingroup$ @YuriNegometyanov Nice. I got it. $\endgroup$
    – River Li
    Jan 26, 2020 at 14:23
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This is too long to fit into a comment. I wanted to ask a question about my proof on this problem. (It might help discover another proof)


This proof has a flaw -- From $AB \ge C$ and $A \ge D$, I wrongly implied that $DB \ge C$.

Is there a way to slightly modify it such that it can prove the statement or is it completely wrong?


Seeing that the inequality is homogeneous (meaning that the transformation $(x, y, z) \mapsto (kx, ky, kz)$ does not change anything), it is natural to impose a constraint on it. So let us assume without the loss of generality that $xyz=1$.

From Cauchy-Schwarz Inequality,

$$([8x^3+5y^3]+[8y^3+5z^3]+[8z^3+5x^3])(\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3})\geqslant (x^2+y^2+z^2)^2$$

Since (By AM-GM) $$[8x^3+5y^3]+[8y^3+5z^3]+[8z^3+5x^3] = 13(x^3+y^3+z^3) \geqslant 13(3 \sqrt[3]{(xyz)^3}) = 13(3)$$

Therefore

$([8x^3+5y^3]+[8y^3+5z^3]+[8z^3+5x^3])(\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3}) \geqslant (13)(3)(\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3}) \geqslant (x^2+y^2+z^2)^2$

Therefore

$$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{(x^2+y^2+z^2)^2}{(13)(3)}$$

Now it reamins to prove that $\frac{(x^2+y^2+z^2)^2}{(13)(3)} \geqslant \frac{x+y+z}{13}$, i.e.

$$(x^2+y^2+z^2)(x^2+y^2+z^2)\geqslant 3(x+y+z)$$

which is straightforward by AM-GM:

Notice that for all $xyz=1$

$$(x - 1)^2 + (y-1)^2 + (z - 1)^2 \ge 0$$ $$x^2 + y^2 + z^2 - 2a - 2b - 2c + 3 \ge 0$$ $$x^2 + y^2 + z^2 \ge -3 + (x + y + z) + (x + y + z)$$

But by AM-GM, $x + y + z \ge 3\sqrt[3]{xyz} = 3$. So, $$x^2 + y^2 + z^2 \ge -3 + 3 + (x + y + z)$$ $$x^2 + y^2 + z^2 \ge x + y + z \ge 3$$

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    $\begingroup$ Your inequality $\sum\limits_{cyc}\frac{x^4}{8x^3+5y^3}\geq\frac{(x^2+y^2+z^2)^2}{(13)(3)}$ it's $\sum\limits_{cyc}\frac{x^4}{8x^3+5y^3}\geq\frac{(x^2+y^2+z^2)^2}{39xyz}$. I think the last inequality is wrong. Try $x=y=1$ and $z=\frac{1}{2}.$ $\endgroup$ Aug 26, 2018 at 1:28
  • $\begingroup$ quoting from the first paragraph: Seeing that the inequality is homogeneous (meaning that the transformation $(x, y, z) \mapsto (kx, ky, kz)$ does not change anything), it is natural to impose a constraint on it. So let us assume without the loss of generality that $xyz=1$. $\endgroup$ Aug 26, 2018 at 2:04
  • $\begingroup$ If you understood it try to check my counterexample. I'll write again: The inequality $\sum\limits_{cyc}\frac{x^4}{8x^3+5y^3}\geq\frac{(x^2+y^2+z^2)^2}{(13)(3)}$ is wrong. Try $x=4$ and $y=z=0.5$. $\endgroup$ Aug 26, 2018 at 6:35
  • $\begingroup$ OK I get it @MichaelRozenberg $\endgroup$ Aug 28, 2018 at 12:47
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$\color{green}{\textbf{Version of 01.04.24. Common way.}}$

$\color{brown}{\;\textbf{1. Cubic in the specific form.}}$

The equation $$\dfrac{3t}{4t^3+d}=c^2,\qquad c>0,\quad c^3d\in[0,1]\tag{1.1}$$ can be solved via the trigonometric form: $$\dfrac{3ct}{4(ct)^3+c^3d}=1,\quad 3ct-4(ct)^3=c^3d,\quad \sin(3\arcsin(ct))=c^3d,$$ $$\mathbf{t_k=\dfrac1c\,sin\;\dfrac{2\pi k+\arcsin(c^3d)}3},\qquad k=0,1,2.\tag{1.2}$$ Since $\;c^3d\in[0,1],\;$ then $\;\arcsin(c^3d)\in\left[0,\dfrac\pi2\right],\;$

$$t_0=\dfrac1c\,\sin\;\dfrac{\arcsin(c^3d)}3\in\left[0,\dfrac1{2c}\right];\tag{1.3}$$ $$t_1=\dfrac1c\,\sin\;\dfrac{2\pi+\arcsin(c^3d)}3 =\dfrac1c\,\sin\;\dfrac{\pi-\arcsin(c^3d)}3\in\left[\dfrac1{2c},\dfrac{\sqrt3}{2c}\right];\tag{1.4}$$ $$t_2=\dfrac1c\,\sin\;\dfrac{4\pi+\arcsin(c^3d)}3=-\dfrac1c\,\sin\;\dfrac{\pi+\arcsin(c^3d)}3\in\left[-\dfrac{\sqrt3}{2c},-\dfrac{1}{c}\right].\tag{1.5}$$

$\color{brown}{\textbf{2. The main areas.}}$

Taking in account the rotational symmetry of the given inequality $$\dfrac{x^4}{8x^3+5y^3}+\dfrac{y^4}{8y^3+5z^3}+\dfrac{z^4}{8z^3+5x^3}\ge\dfrac{x+y+z}{13},\tag{2.1}$$ we can WLOG assume $\;x\;$ the greatest one of the unknowns $\;\{x,y,z\}.\;$

Then it suffices to consider two main cases (areas):

$A.\quad 0\le z\le y\le x.$

$B.\quad0\le y\le z\le x.$

$\color{brown}{\textbf{3. Edges.}}$

$\color{teal}{\textbf{3a. Edges,case a).}}$

Let $\;\color{teal}{\mathbf{\;z=0,\;0<y\le x.\;}}$ Then inequality ($2.1$) takes a form of $$\dfrac{x^4}{8x^3+5y^3}+\dfrac{y^4}{8y^3}\ge\dfrac{x+y}{13}.$$ Substitution $u=\dfrac yx\in[0,1]\;$ presents it in the form of $\;L(u)\ge0,\;$ where $$L(u)=\dfrac1{8+5u^3}+\dfrac18u-\dfrac{1+u}{13} =\dfrac{104+13u(8+5u^3)-8(1+u)(8+5u^3)}{104(8+5u^3)}$$ $$=\dfrac{40+40u-40u^3+25u^4}{104(8+5u^3)} =\dfrac5{104}\dfrac{8+8u(1-u^2)+5u^4}{8+5u^3}\ge0.$$

The case is proven.

$\color{teal}{\textbf{3b. Edges, case b).}}$

Let $\color{teal}{\mathbf{\;y=0,\;0<z\le x.\;}}$ Then inequality ($2.1$) takes a form of $$\dfrac{x^4}{8x^3}+\dfrac{z^4}{8z^3+5x^3}\ge\dfrac{x+z}{13}.$$ Substitution $\;u=\dfrac zx\in[0,1]\;$ presents it in the form of $\;L(u)\ge0,\;$ where $$L(u)=\dfrac18+\dfrac{u^4}{5+8u^3}-\dfrac{1+u}{13} =\dfrac{13(5+8u^3)+104u^4-8(1+u)(5+8u^3)}{104(5+8u^3)}$$ $$=\dfrac{25-40u+40u^3+40u^4}{104(5+8u^3)} =\dfrac{15u^4+17(1-u^2)^2+2(2-5u-2u^2)^2}{104(5+8u^3)}\ge0.$$ The case is proven.

Therefore,inequality ($2.1$) $\color{green}{\textbf{is proven at the bounds of the given area.}}$

$\color{brown}{\textbf{4. Substitutions.}}$

In the case "A", $\;\color{teal}{\mathbf{0<z\le y\le x.}}$

Substitutions $$u=\dfrac yx\in[0,1],\quad v=\dfrac zy\in[0,1],\quad w=\dfrac xz=\dfrac1{uv}\in[1,\infty),\tag{4.1}$$ lead to presentation of the given inequality in the form of $$\dfrac{1}{8+5u^3}-\dfrac1{13}+u\left(\dfrac{1}{8+5v^3}-\dfrac1{13}\right)+uv\left(\dfrac1{8+5w^3}-\dfrac1{13}\right)\ge 0.\tag{4.2}$$

In the case "B", $\;\color{teal}{\mathbf{0<y\le z\le x.}}$

Substitutions $$u=\dfrac zy\in[1,\infty],\quad v=\dfrac xz\in[1,\infty],\quad w=\dfrac yx=\dfrac1{uv}\in[0,1],\tag{4.3}$$ lead to presentation of the given inequality in the form of $$\dfrac{1}{8+5w^3}-\dfrac1{13}+\dfrac1{uv}\left(\dfrac{1}{8+5u^3}-\dfrac1{13}\right)+\dfrac1v\,\left(\dfrac1{8+5v^3}-\dfrac1{13}\right)\ge0.\tag{4.4}$$ Both ($\text{4.2}$) and ($\text{4.4}$) can be written in the form of $$\dfrac1{uv}\,\dfrac{1-u^3}{8+5u^3}+\dfrac1v\,\dfrac{1-v^3}{8+5v^3}+\dfrac{1-w^3}{8+5w^3}\ge 0.\tag{4.5}$$

At the same time, the constraints ($\text{4.1}$) and ($\text{4.3}$) are specific ones.

Let $$\begin{cases} P(t)=\dfrac{1-t^3}{8+5t^3},\\[4pt] Q(t)=-P'(t)=\dfrac{39t^2}{(8+5t^3)^2},\\[4pt] R(t)=P(t)+tQ(t)=\dfrac{8+36t^3-5t^6}{(8+5t^3)^2},\tag{4.6} \end{cases}$$ then (${4.5}$) takes the form of $\;f(u,v)\ge0,\;$ where $$f(u,v)=P\left(\dfrac1{uv}\right)+\dfrac1v\left(P(v)+\dfrac1u\,P(u)\right).\tag{4.7}$$ $\color{brown}{\textbf{5. Inner stationary points.}}$

$\color{teal}{\textbf{5.1. Algebraic system.}}$

Stationary points can be defined from the system $\;f'_u=f'_v=0,\;$ or $$\begin{cases} P'\left(\dfrac1{uv}\right)\left(-\dfrac1{u^2v}\right)+\dfrac1v\left(-\dfrac1{u^2}\right)P(u)+\dfrac1{uv}\,P'(u)=0\\[4pt] P'\left(\dfrac1{uv}\right)\left(-\dfrac1{uv^2}\right)-\dfrac1{v^2}\left(P(v)+\dfrac1uP(u)\right)+\dfrac1vP'(v)=0, \end{cases}$$ $$\begin{cases} Q(w)-P(u)-uQ(u)=0\\[4pt] Q(w)-P(u)-uP(v)-uvQ(v)=0\\[4pt] uvw=1 \end{cases}\Rightarrow \begin{cases} R(u)=Q(w)\\[4pt] R(v)=Q(u)\\[4pt] uvw=1 \end{cases}\tag{5.1}$$

$\color{teal}{\textbf{5.2. The inequality transformation.}}$

Taking in account the equalities $(4.6),(4.7),(5.1)$ in the form of $$P(t)=R(t)-tQ(t),\quad R(u)=Q(w),\quad R(v)=Q(u),\quad uvw=1,$$ the inequality $$wf(u,v,w)=P(w)+w(P(u)+uP(v))\ge0$$ can be presented in a form of $$R(w)-wQ(w)+w\big(R(u)-uQ(u)+uR(v)-uvQ(v)\big)\ge0,$$ $$R(w)+w\big(R(u)-Q(w)\big)+uw\big(R(v)-Q(u)\big)-uvwQ(v)\ge0,$$ or $$R(w)-Q(v)\ge0.\tag{5.2}$$

$\color{brown}{\textbf{6. Explicit form of the inverse functions.}}$

$\color{teal}{\textbf{6.1. Inversion of $Q(t).$}}$

Function $Q(t)$ can be presented in the form of $$Q(t)= \dfrac{39t^2}{(8+5t^3)^2}=\dfrac{208}{75}\left(\dfrac{3t}{\dfrac{32}5+4t^3}\right)^2.\tag{6.1}$$ Q(t)

Then

  • $Q(0)=0;$
  • $\;\lim\limits_{t\to\infty}Q(t)=0;$
  • $\;\dfrac{\text d}{\text dt}\dfrac{3t}{8+5t^3}=\dfrac{6(4-5t^3)}{(8+5t^3)^2};\;$
  • $\max Q(t)=Q_m=\dfrac{13\sqrt[\large3]{10}}{120}\approx0.233397091\;$at$\;t_m=\sqrt[\large3]{\dfrac45}\approx0.928317767.$
  • $Q(t)\;$ increases at $\;t\in[0,t_m];$
  • $Q(t)\;$ decreases at $\;t\in[t_m,\infty].$

Taking in account $(1.1)-(1.2),$ inverse function $\;Q^{-1}(q)\;$ finally is $$\theta(k,q)=2\sqrt[\large4]{\dfrac{13}{75q}} \sin\left(\dfrac13\left(2\pi k+\arcsin\left(\dfrac{4}5\left(\dfrac{75q}{13}\right)^{\large\frac34}\right)\right)\right),\quad k=0,1.\tag{6.2}$$

Note that, in accordance with $(1.5),\quad\theta(2,q)<0.$

Therefore, there are two branches of the inverse function $Q^{-1}(q):$

  • "zero" branch $\;\theta(0,q),\quad q\in\left(0,Q_m\right],\quad\theta(0,q)\in[0,t_m],\;$ and
  • "first" branch $\;\theta(1,q),\quad q\in\left(0,Q_m\right],\quad \theta(1,q)\in[t_m,\infty].$

Easily to check,that

  • $Q(\theta(k,q))=q,\quad q\in(0,\infty);$
  • $\theta(0,Q(t))=t,\quad t\in[0,t_m];$
  • $\theta(1,Q(t))=t,\quad t\in[t_m,\infty].$

$\color{teal}{\textbf{6.2. Inversion of $R(t).$}}$

Function $R(t)$ can be presented in the form of $$R(t)=\dfrac{8+36t^3-5t^6}{(8+5t^3)^2}=\dfrac7{30}-\dfrac{13}{30}\left(\dfrac{12}{8+5t^3}-1\right)^2.\tag{6.3}$$ r(t)

Easily to see, that

  • $R(0)=\dfrac18=0.125;$
  • the positive root of $R(t)$ is $\;t_h=\sqrt[\large3]{\dfrac25\,\left(9+\sqrt{91}\right)} \approx1.950077297;$
  • $\max R(t)=R_m=\dfrac7{30}\approx0.233333333\;$ at $\;t_m=\sqrt[\large3]{\dfrac45}\approx0.928317767;$
  • $R(t)\;$ increases at $\;t\in[0,t_m];$
  • $R(t)\;$ decreases at $\;t\in[t_m,t_h];$
  • $R(t)\;$ is negative at $\;t\in(t_h,\infty).$

Finally, the inverse function is $$\rho(s,r)=2\left(\dfrac{0.3}{1+s\sqrt{\dfrac{7-30r}{13}}}-0.2\right)^{\large\frac13},\tag{6.4}$$ where $\;s=\pm1.$

Accordingly,there are two branches of the inverse function $R^{-1}(r):$

  • "plus"-branch $\;\rho(1,r),\quad r\in\left[\dfrac18,\dfrac7{30}\right], \quad\rho(1,r)\in[0,t_m],\;$ and
  • "minus"-branch $\;\rho(-1,r),\quad r\in\left[0,\dfrac7{30}\right], \quad\rho(-1,r)\in[t_m,t_h].$

Easily to check, that

  • $R(\rho(\pm1,r))=r,\quad r\in[0,\infty);$
  • $\rho(1,R(t))=t,\quad t\in[0,t_m];$
  • $\rho(-1,R(t))=t,\quad t\in[t_m,t_h].$

$\color{teal}{\textbf{6.3. Explicit expressions.}}$

Since $$Q(t)\in(0,Q_m),\quad R(t)\in(0,R_m),\quad R_m=\dfrac7{30}<Q_m,$$ then from $(5.1)$ should $$Q(t)\in\left(0,\dfrac7{30}\right],\quad t\in\left((0,t_0]\cup[t_1,t_h]\right),\tag{6.5}$$ where $$\begin{cases} t_0=\theta\left(0,\dfrac7{30}\right]\approx0.917509714,\\[4pt] t_m=\sqrt[\large3]{\dfrac45}\approx0.928317767,\\[4pt] t_1=\theta\left(1,\dfrac7{30}\right]\approx0.939210366,\\[4pt] t_h=\rho(-1,0)\approx1.950077297.\end{cases} \tag{6.6}$$

From $\;(5.1)\;R(u)=Q(w)\;$ should $\;u=\rho(s,Q(w))=g(s,w),\;$ where $$g(s,t)=2\left(\dfrac{0.3}{1+s\sqrt{\dfrac7{13}-\dfrac{90t^2}{(8+5t^3)^2}}}-0.2\right)^{\large\frac13},\quad s=-1,1.\tag{6.7}$$ Also, from $\;(5.1)\;R(v)=Q(u)\;$ should $v=g(s,u).$

On the other hand, from $\;(5.1)\;R(u)=Q(w)\;$ should $w=\theta(k,R(u))=h(k,u),\;$ where $$h(k,t)=2\sqrt[\large4]{\dfrac{13}{75R(t)}} \sin\left(\dfrac13\left(2\pi k+\arcsin\left(\dfrac{4}5\left(\dfrac{75R(t)}{13}\right) ^{\large\frac34}\right)\right)\right),\quad k=0,1.\tag{6.8}$$ Also,from $\;(5.1)\;R(v)=Q(u)\;$ should $u=h(k,v).$

At the same time,in accordance with $(5.1)$ should $$ug(s,u)h(k,u)=1.\tag{6.9}$$

$\color{brown}{\textbf{7. Proof, case A).}}$

In accordance with $(4.1),\quad u,v\in(0,1],\quad w\in[1,t_h],\quad $ and then should $$\;R(u)\in\left(\dfrac18,\dfrac7{30}\right],\tag{7.1}$$ $$u w\ge1.\tag{7.2}$$

$\color{teal}{\textbf{7.1. Case A,$\;u\in[0,t_0],\;k=0.$}}$

Taking in account $(1.3),(7.1),(7.2),$ easily to get $$w=h(0,u)=\dfrac1{c(u)}\,\sin\left(\dfrac13\,\arcsin\left(\dfrac{32}5\,c^3(u)\right)\right),\tag{7.3}$$ where $$с(u)=\sqrt[\large4]{\dfrac{75R(u)}{208}}\in\left[\sqrt[\large4]{\dfrac{75}{208}\,\dfrac18},\sqrt[\large4]{\dfrac{75}{208}\dfrac7{30}}\right]\subset[0.460762350,0.538571886],\tag{7.4}$$ $$\dfrac1{c(u)}\le\sqrt[\large4]{\dfrac{208\cdot8}{75}}<2.170316217,$$ $$\dfrac{32c^3(u)}5\le\dfrac{32}5\left(\dfrac{75}{208}\cdot\dfrac7{30}\right)^{\large\frac34}<0.999795113,$$ $$uh(0,u)<t_0\cdot\dfrac1{c(u)}\cdot\dfrac12<1.$$ and this contradicts with $(7.2).$

The plot of $\;uh(0,u)\;$ is shown below.

Plot of uh(0,u)

$\color{teal}{\textbf{7.2. Case A,$\;u\in[0,t_0],\;k=1.$}}$

Taking in account $(1.4),$ one can get $$\begin{align} &w=h(1,u)=\sqrt[\large4]{\dfrac{208}{75}\dfrac{(8+5u^3)^2}{8+36u^3-5u^6}}\\[4pt]&\times\sin\left(\dfrac13\left(\pi-\arcsin\left(\dfrac45\left(\dfrac{75}{13}\dfrac{8+36u^3-5u^6}{(8+5u^3)^2}\right)^{\large\frac34} \right)\right)\right), \end{align}\tag{7.5}$$ wherein both of multipliers are decreasing functions of $u.$

Let $\;H(t)=\dfrac1{h(1,t)}.$

Taking in account $(7.2)$ in the form of $\;u\ge\dfrac1w,\;$ one can get:

  • for $\;u\in[0, H(0)]\qquad\quad\;\;\, u h(1,u) \le 1,\qquad H(0)\approx 0.629192687,$
  • for $\;u\in[H(0), H(H(0))]\quad u h(1,u) \le 1,\qquad H(H(0))\approx 0.805791121,$
  • for $\;u\in[H(H(0)),t_0]\qquad\; u h(1,u) \le 1,\qquad t_0\approx 0.917509714.$

Therefore,

  • for $\;u\in[0,t_0]\quad u h(1,u) \le 1.$

The plot of $\;uh(1,u)\;$ is shown below.

Plot of uW(1,u)

$\color{teal}{\textbf{7.3. Case A,$\;u\in(t_0,t_1).$}}$

If $\;u\in(t_0,t_1),\;$ then $\;Q(u)\in\left(\dfrac7{30},\dfrac{13\sqrt[\large3]{10}}{120}\right).$

At the same time, from $(5.1)$ should the equation $$Q(u)=R(v),\quad R(v)\in\left[0,\dfrac7{30}\right],$$ which has not real solutions.

$\color{teal}{\textbf{7.4. Case A,$\;u\in[t_1,1].$}}$

Since $\;R(u)\in\left(\dfrac3{13},\dfrac7{30}\right],\;$ then

$\;w=h(0,u)\in[0.860147050,0.917509715],\;$ in contradiction with the conditions $(4.1).$

$\;(w=h(1,u)\in[0.943496443,1])\wedge(w\in[1,\infty)),$ with the solution $\;\color{green}{\mathbf{\{u,v,w\}=\{1,1,1\}}},\;$ which corresponds with the inequality $(4.5).$

Therefore, the inequality $(4.5)\;\color{green}{\textbf{is proven in the case A}}.$

$\color{brown}{\textbf{8. Proof,case B,$\;u\in[1,t_h]$.}}$

$\color{teal}{\textbf{8.1. Preliminary notes.}}$

Taking in account $(4.3),$ should $\;u\in(1,t_h),\quad v\in(1,\infty),\quad w\in(0,1),\quad t_h=\rho(-1,0).$

Then $$\color{teal}{\mathbf{v=g(-1,u),\quad w=h(0,u)}}.\tag{8.1}$$

Graphic analysis of the functions $\;h(0,u)\;$ (orange line) and $\;u g(-1,u)\;$ (green line)

ug(-1,u), h(0,u)

show, that both of them allow accurate linear approximations at the interval $\;u\in(1, t_h):$ $$h(0,u)\approx 0.860147-0.796(u-1)\pm 0.0133,\quad ug(-1,u)\approx 1+2.3(u-1)\pm 0.04.$$ So their production can be approximated via quadratic function.

Therefore, the function $B(u)= u g(-1,u) h(0,u)$ is unimodal one in the interval $u\in(1,t_h).$

$\color{teal}{\textbf{8.2. Proof.}}$

From $(8.1),(6.7),(6.8)$ should that $$B(u)=ug(-1,u)h(0,u)=1\tag{8.2}$$ in the stationary points of the $f(u,v).$

Since $B(u)$ has the single maximum at $u\in(1,t_h),$ then equation $(8.2)$ can not have more than two roots in this interval.

Let us calculate the table $$\hspace{-32mu}\begin{vmatrix} u & v=g(-1,u) & w=\theta(0,R(u)) & uvw & R(w) & Q(v)\\ \!1.25&\!1.228103035&\!0.650433454&\!0.998499124&\!0.199388943&\!0.197417010\\ \!1.26&\!1.236402046&\!0.642696247&\!1.001235004&\!0.197754855&\!0.195782736\\ \!1.47&\!1.395850784&\!0.487833698&\!1.000986283&\!0.164510886&\!0.162892269\\ \!1.48&\!1.402726182&\!0.480700601&\!0.997951152&\!0.163086876&\!0.161467543\tag{8.3} \end{vmatrix}$$

Easily to see that there are two solutions of $(7.4)$,which are situated at the intervals $(1.25,1.26)$ and $(1.47,1.48)$ accordingly. At the same time, the functions $R(w)$ and $Q(v)$ are monotonic in these intervals. In particular, in the interval $(1.25,1.26)\quad R(w)>0.1977\;$ and $\;Q(v)<0.1975.$ This means that the inequality $(7.2)$ is satisfied at the interval $\;u\in(1.25,1.26),$ including the stationary point.

Similarly, at the interval $(1.47,1.48)$ easily to get $\;R(w)>0.1630>0.1629>Q(w).$ This means that the inequality $(7.2)$ is satisfied at the interval $\;u\in(1.47,1.48),$ including the stationary point.

Plot of the equality $uvw=1$ see below.

Plotuvw=1

Therefore, the inequality $(4.5)$ $\color{green}{\textbf{is satisfied in the case B}}.$

$\color{green}{\textbf{Proved!!!}}$

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We want to prove the inequality $$ \sum_{cyc}\frac{x^4}{8x^3+5y^3}\geq \frac{\sum_{cyc}x}{13}\tag 1 $$ where $x,y,z>0$.

Set $$ \Pi(x,y,z):=\sum_{cyc}\frac{x^4}{8x^3+5y^3}-\frac{x+y+z}{13} $$ Then easily $$ \Pi(x,y,z)= \sum_{cyc}\left(x\frac{x^3}{8x^3+5y^3}-\frac{x}{13}\right)=\sum_{cyc}x\frac{13x^3-8x^3-5y^3}{13(8x^3+5y^3)}= $$ $$ =\frac{5}{13}\sum_{cyc}x\frac{x^3-y^3}{8x^3+5y^3}. $$ Also (ecxept from the symetry) $\Pi(x,y,z)$ is homogeneous of degree 1. Hence $\Pi(\lambda x,\lambda y,\lambda z)=\lambda \Pi(x,y,z)$.

The case $x\geq y\geq z$

If $x\geq y\geq z$, we can set $\lambda=1/z$, then $$ \Pi(x,y,z)=\lambda^{-1}\Pi(\lambda x,\lambda y,\lambda z)=z^{-1}\Pi\left(\frac{x}{z},\frac{y}{z},1\right). $$ Set $a=\frac{x}{z}$, $b=\frac{y}{z}$, then also $a\geq b\geq 1$ and $$ \Pi(x,y,z)=z\Pi(a,b,1)= $$ $$ =z\left(\frac{1-a^3}{8+5a^3}+\frac{b(-1+b^3)}{5+8b^3}+\frac{a^4-ab^3}{8a^3+5b^3}\right)\textrm{, }a\geq b\geq 1\tag 2 $$ Set now $$ f(x,y)=\frac{x}{8x^3+5 y^3}. $$ Then (2) becomes $$ a^3\left(\frac{a}{8a^3+5b^3}-\frac{1}{8+5a^3}\right)+b^3\left(\frac{b}{5+8b^3}-\frac{a}{8a^3+5b^3}\right)- $$ $$ -\left(\frac{b}{8b^3+5}-\frac{1}{8+5a^3}\right)\geq 0 $$ Hence equivalent we must show that $$ a^3(f(a,b)-f(1,a))+b^3(f(b,1)-f(a,b))-(f(b,1)-f(1,a))\geq 0\Leftrightarrow $$ $$ a^3f(a,b)+f(1,a)(1-a^3)\geq b^3f(a,b)+f(b,1)(1-b^3)\Leftrightarrow $$ $$ f(1,a)(a^3-1)-f(b,1)(b^3-1)\leq (a^3-b^3)f(a,b)\tag 3 $$ But when $a\geq b\geq 1$, setting $k=a^3-1$ and $l=b^3-1$, we have $k\geq l\geq 0$. Then (3) becomes $$ k\frac{1}{5a^3+8}-l\frac{b}{8b^3+5}\leq (k-l)\frac{a}{8a^3+5b^3}\Leftrightarrow $$ $$ \frac{k}{5k+13}-\frac{lb}{8l+13}\leq (k-l)\frac{a}{8k+5l+13}\Leftrightarrow $$ $$ \frac{k}{k-l}\frac{1}{x_1}-\frac{l}{k-l}\frac{b}{y_1}\leq \frac{a}{8(x_1-13)/5+5(y_1-13)/8+13} $$ Or equivalent (we set $x_1=5k+13$, $y_1=8l+13$) if: $$ \small f_0(x_1,y_1):=5ky_1^2/8-8blx_1^2/5+x_1y_1(al-ak+8k/5-5bl/8)-637ky_1/40+637bl x_1/40\leq 0.\normalsize $$ This last inequality is true, as one can see using $1\leq b\leq a$ and $k\geq l\geq 0$, $k=a^3-1$, $l=b^3-1$, $x_1\geq 13$, $y_1\geq 13$. Actualy it holds $$ f_0(x_1,y_1)<f_0(x_0,y_0)\textrm{, }\forall (x_1,y_1)\in D=(x_0,+\infty)\times(y_0,+\infty) $$ where $$\small x_0=-\frac{637 k (8 (5 a-8) k-5 l (8 a+5 b))}{-80 k l (8 a (5 a-8)-5 (5 a+8) b)+25 l^2 (8 a-5 b)^2+64 (8-5 a)^2 k^2}\normalsize $$ $$\small y_0=\frac{637 b l (5 l (5 b-8 a)+8 (5 a+8) k)}{-80 k l (8 a (5 a-8)-5 (5 a+8) b)+25 l^2 (8 a-5 b)^2+64 (8-5 a)^2 k^2}\normalsize $$ are the points such that $\partial_xf_0(x_0,y_0)=\partial_yf_0(x_0,y_0)=0$ and $$\small f_0(x_0,y_0)=-\frac{405769 a b k l (k-l)}{-80 k l (8 a (5 a-8)-5 (5 a+8) b)+25 l^2 (8 a-5 b)^2+64 (8-5 a)^2 k^2}\leq 0\normalsize $$ and $64 (8 - 5 a)^2 k^2 - 80 (8 a (-8 + 5 a) - 5 (8 + 5 a) b) k l + 25 (8 a - 5 b)^2 l^2>0$, $a>1$, $a\geq b\geq 1$. But when $x_1\geq 13$ and $y_1\geq 13$, we have $f_0(x_1,y_1)\leq 0$.

About the case $z\geq y\geq x$

If $z\geq y\geq x$, then we have $\Pi(\lambda x,\lambda y,\lambda z)=\lambda \Pi(x,y,z)$. Set $\lambda=1/x$ and $a=z/x$, $b=y/x$, $a\geq b\geq 1$. Then $$ \Pi(x,y,z)=x^{-1}\Pi(1,b,a)= $$
$$ b^3\left(\frac{b}{8b^3+5a^3}-\frac{1}{8+5b^3}\right)+a^3\left(\frac{a}{5+8a^3}-\frac{b}{8b^3+5a^3}\right)- $$ $$ -\left(\frac{a}{8a^3+5}-\frac{1}{8+5b^3}\right)\geq 0\tag 5 $$ Set $$ f(x,y)=\frac{x}{8x^3+5y^3}, $$ then (5) becomes $$ b^3(f(b,a)-f(1,b))+a^3(f(a,1)-f(b,a))-f(a,1)+f(1,b)\geq 0\Leftrightarrow $$ $$ (a^3-1)f(a,1)-(b^3-1)f(1,b)\geq (a^3-b^3)f(b,a). $$ Set now $k=a^3-1$ and $l=b^3-1$. Then $k\geq l\geq 0$. Then $$ k\frac{a}{8 a^3+5}-l\frac{1}{8+5 b^3}\geq (k-l)\frac{b}{8b^3+5a^3}\Leftrightarrow $$ $$ \frac{ka}{8k+13}-\frac{l}{5 l+13}\geq (k-l)\frac{b}{8k+5l+13}\Leftrightarrow $$ $$ \frac{ka}{k-l}\frac{1}{x_1}-\frac{l}{k-l}\frac{1}{y_1}\geq \frac{b}{x_1+y_1-13} $$ Or equivalent ($x_1=8k+13$, $y_1=5l+13$): $$ f_0(x_1,y_1):=aky_1^2-lx_1^2+x_1y_1(bl-l-bk+ak)-13aky_1+13l x_1\geq 0. $$ This last inequality is true and as one can see using $1\leq b\leq a$ and $k\geq l\geq 0$, $k=a^3-1$, $l=b^3-1$, $x_1\geq 13$, $y_1\geq 13$. Actualy holds $$ f_0(x_1,y_1)>f_0(x_0,y_0)\textrm{, }\forall (x_1,y_1)\in D=(x_0,+\infty)\times(y_0,+\infty) $$ where $$ x_0=\frac{13ak(ak+l+b(l-k))}{a^2k^2+(b(k-l)+l)^2+2ak(l+b(l-k))} $$ $$ y_0=\frac{13l((a+b)k+l-lb)}{a^2k^2+(b(k-l)+l)^2+2ak(l+b(l-k))} $$ are the points such that $\partial_xf(x_0,y_0)=\partial_yf_0(x_0,y_0)=0$ and $$ f_0(x_0,y_0)=\frac{169abkl(l-k)}{a^2k^2+(b(k-l)+l)^2+2ak(l+b(l-k))}\leq 0 $$ and $a^2k^2+(b(k-l)+l)^2+2ak(l+b(l-k))>0$, $a>1$, $a\geq b\geq 1$. But when $x_1\geq 13$ and $y_1\geq 13$, we have $f_0(x_1,y_1)\geq 0$. QED

$\endgroup$
14
  • $\begingroup$ You consider the case $x\ge y\ge z$, right? How about the left case? $\endgroup$
    – River Li
    Dec 25, 2020 at 1:07
  • $\begingroup$ @RiverLi. Thanks for the note. Can you check it now please? $\endgroup$ Dec 26, 2020 at 23:35
  • $\begingroup$ 1) It should be $\Pi(x,y,z)=\lambda^{-1}\Pi(\lambda x,\lambda y,\lambda z)=z\Pi\left(\frac{x}{z},\frac{y}{z},1\right)$. $\endgroup$
    – River Li
    Dec 27, 2020 at 4:26
  • $\begingroup$ 2) In (2), is should be $\Pi(x,y,z)=z\Pi(a,b,1)= \frac{5}{13}z (\frac{1-a^3}{8+5a^3}+\frac{b(-1+b^3)}{5+8b^3}+\frac{a^4-ab^3}{8a^3+5b^3})$ $\endgroup$
    – River Li
    Dec 27, 2020 at 4:26
  • $\begingroup$ 3) In (3), $f(1,a)(1-a^3)\geq f(b,1)(1-b^3)$ is not true (check $a = 2$, $b = 1$). $\endgroup$
    – River Li
    Dec 27, 2020 at 4:40
-1
$\begingroup$

Some explanations for the simple case $x\geq y\ge z$:

Define the functions :

$$h\left(x\right)=\frac{2}{1+x^{3}}+\frac{x^{3}+x}{\frac{4}{3}+x^{3}}+\frac{\frac{1}{3}x}{x+2}$$

$$g\left(x\right)=\frac{13}{8+5x^{3}}$$

For $x>0$ one can show that :

$$g''(x)-h''(x)>0$$

So using Jensen's inequality we get a lower bound .

Remains to show that :

$$a(x,y,z)=\frac{x}{x+y+z}h\left(\frac{y}{x}\right)+\frac{y}{x+y+z}h\left(\frac{z}{y}\right)+\frac{z}{x+y+z}h\left(\frac{x}{z}\right)-1+g\left(1\right)-h\left(1\right)\geq 0$$

Now using Buffalo's way in $a(x+y+z,x+y,x)$ allow us to conclude .

We have in fact :

$$a(x+y+z,x+y,x)=(384z¹⁵ y⁷ + 2880z¹⁵ y⁶ x + 9216z¹⁵ y⁵ x² + 16992z¹⁵ y⁴ x³ + 20304z¹⁵ y³ x⁴ + 15984z¹⁵ y² x⁵ + 7824z¹⁵ y x⁶ + 2016z¹⁵ x⁷ + 17424z¹⁴ y⁸ + 152280z¹⁴ y⁷ x + 579168z¹⁴ y⁶ x² + 1268244z¹⁴ y⁵ x³ + 1772010z¹⁴ y⁴ x⁴ + 1633392z¹⁴ y³ x⁵ + 977040z¹⁴ y² x⁶ + 347490z¹⁴ y x⁷ + 53172z¹⁴ x⁸ + 208344z¹³ y⁹ + 2062764z¹³ y⁸ x + 9027612z¹³ y⁷ x² + 23079510z¹³ y⁶ x³ + 38250549z¹³ y⁵ x⁴ + 42851619z¹³ y⁴ x⁵ + 32551944z¹³ y³ x⁶ + 16139979z¹³ y² x⁷ + 4636053z¹³ y x⁸ + 531846z¹³ x⁹ + 1283160z¹² y¹⁰ + 14188068z¹² y⁹ x + 70192728z¹² y⁸ x² + 205499258z¹² y⁷ x³ + 396004119z¹² y⁶ x⁴ + 526697346z¹² y⁵ x⁵ + 490587345z¹² y⁴ x⁶ + 315636651z¹² y³ x⁷ + 132957702z¹² y² x⁸ + 31969981z¹² y x⁹ + 2971122z¹² x¹⁰ + 4971312z¹¹ y¹¹ + 60733188z¹¹ y¹⁰ x + 335178450z¹¹ y⁹ x² + 1106426454z¹¹ y⁸ x³ + 2434804950z¹¹ y⁷ x⁴ + 3759909216z¹¹ y⁶ x⁵ + 4163358042z¹¹ y⁵ x⁶ + 3304149732z¹¹ y⁴ x⁷ + 1833331446z¹¹ y³ x⁸ + 667237116z¹¹ y² x⁹ + 137638266z¹¹ y x¹⁰ + 10832808z¹¹ x¹¹ + 13298856z¹⁰ y¹² + 177961950z¹⁰ y¹¹ x + 1084146345z¹⁰ y¹⁰ x² + 3985363576z¹⁰ y⁹ x³ + 9869364894z¹⁰ y⁸ x⁴ + 17379228279z¹⁰ y⁷ x⁵ + 22339972161z¹⁰ y⁶ x⁶ + 21120138531z¹⁰ y⁵ x⁷ + 14540414355z¹⁰ y⁴ x⁸ + 7058059577z¹⁰ y³ x⁹ + 2252276967z¹⁰ y² x¹⁰ + 406736841z¹⁰ y x¹¹ + 28000728z¹⁰ x¹² + 25847784z⁹ y¹³ + 376182900z⁹ y¹² x + 2508149808z⁹ y¹¹ x² + 10164572120z⁹ y¹⁰ x³ + 27991403643z⁹ y⁹ x⁴ + 55403735649z⁹ y⁸ x⁵ + 81174314926z⁹ y⁷ x⁶ + 89154075117z⁹ y⁶ x⁷ + 73315454478z⁹ y⁵ x⁸ + 44417254351z⁹ y⁴ x⁹ + 19100032290z⁹ y³ x¹⁰ + 5417205333z⁹ y² x¹¹ + 871662107z⁹ y x¹² + 53844294z⁹ x¹³ + 37566720z⁸ y¹⁴ + 591122286z⁸ y¹³ x + 4283190279z⁸ y¹² x² + 18978713280z⁸ y¹¹ x³ + 57556578552z⁸ y¹⁰ x⁴ + 126569962581z⁸ y⁹ x⁵ + 208342370412z⁸ y⁸ x⁶ + 260878006686z⁸ y⁷ x⁷ + 249573066114z⁸ y⁶ x⁸ + 181161461376z⁸ y⁵ x⁹ + 97759315011z⁸ y⁴ x¹⁰ + 37666529904z⁸ y³ x¹¹ + 9613721448z⁸ y² x¹² + 1399093395z⁸ y x¹³ + 79092636z⁸ x¹⁴ + 41415984z⁷ y¹⁵ + 701192430z⁷ y¹⁴ x + 5489681229z⁷ y¹³ x² + 26414947705z⁷ y¹² x³ + 87517530588z⁷ y¹¹ x⁴ + 211799778048z⁷ y¹⁰ x⁵ + 387184982171z⁷ y⁹ x⁶ + 544718091222z⁷ y⁸ x⁷ + 594555654477z⁷ y⁷ x⁸ + 502941420053z⁷ y⁶ x⁹ + 326174555253z⁷ y⁵ x¹⁰ + 158433882981z⁷ y⁴ x¹¹ + 55253699785z⁷ y³ x¹² + 12826926003z⁷ y² x¹³ + 1708355601z⁷ y x¹⁴ + 89637030z⁷ x¹⁵ + 34720080z⁶ y¹⁶ + 630057006z⁶ y¹⁵ x + 5304841911z⁶ y¹⁴ x² + 27564533944z⁶ y¹³ x³ + 99116893383z⁶ y¹² x⁴ + 261925497270z⁶ y¹¹ x⁵ + 526781771279z⁶ y¹⁰ x⁶ + 823053262677z⁶ y⁹ x⁷ + 1009782652953z⁶ y⁸ x⁸ + 975564801912z⁶ y⁷ x⁹ + 738623531838z⁶ y⁶ x¹⁰ + 432336411336z⁶ y⁵ x¹¹ + 190731895504z⁶ y⁴ x¹² + 60716879538z⁶ y³ x¹³ + 12925101726z⁶ y² x¹⁴ + 1588005317z⁶ y x¹⁵ + 78040326z⁶ x¹⁶ + 21919608z⁵ y¹⁷ + 425170674z⁵ y¹⁶ x + 3835887885z⁵ y¹⁵ x² + 21427680459z⁵ y¹⁴ x³ + 83173337562z⁵ y¹³ x⁴ + 238463234604z⁵ y¹² x⁵ + 523581046341z⁵ y¹¹ x⁶ + 899993685876z⁵ y¹⁰ x⁷ + 1226601376371z⁵ y⁹ x⁸ + 1332872926572z⁵ y⁸ x⁹ + 1153802001282z⁵ y⁷ x¹⁰ + 789677299992z⁵ y⁶ x¹¹ + 420642512409z⁵ y⁵ x¹² + 169737375000z⁵ y⁴ x¹³ + 49613151276z⁵ y³ x¹⁴ + 9728891787z⁵ y² x¹⁵ + 1106165466z⁵ y x¹⁶ + 51133236z⁵ x¹⁷ + 10164360z⁴ y¹⁸ + 210442770z⁴ y¹⁷ x + 2029540995z⁴ y¹⁶ x² + 12147507662z⁴ y¹⁵ x³ + 50683080903z⁴ y¹⁴ x⁴ + 156835746669z⁴ y¹³ x⁵ + 373573760956z⁴ y¹² x⁶ + 701067124917z⁴ y¹¹ x⁷ + 1051426606638z⁴ y¹⁰ x⁸ + 1269761155782z⁴ y⁹ x⁹ + 1237154359086z⁴ y⁸ x¹⁰ + 968999923944z⁴ y⁷ x¹¹ + 604190085863z⁴ y⁶ x¹² + 294655270920z⁴ y⁵ x¹³ + 109205348349z⁴ y⁴ x¹⁴ + 29362605619z⁴ y³ x¹⁵ + 5297785974z⁴ y² x¹⁶ + 555034197z⁴ y x¹⁷ + 24079356z⁴ x¹⁸ + 3298656z³ y¹⁹ + 72955782z³ y¹⁸ x + 751596237z³ y¹⁷ x² + 4810891429z³ y¹⁶ x³ + 21511743804z³ y¹⁵ x⁴ + 71559881460z³ y¹⁴ x⁵ + 183985237732z³ y¹³ x⁶ + 374624922801z³ y¹² x⁷ + 613551007287z³ y¹¹ x⁸ + 815662822327z³ y¹⁰ x⁹ + 883653119070z³ y⁹ x¹⁰ + 779429014659z³ y⁸ x¹¹ + 556423335640z³ y⁷ x¹² + 317665392660z³ y⁶ x¹³ + 142179626826z³ y⁵ x¹⁴ + 48350388565z³ y⁴ x¹⁵ + 11889412725z³ y³ x¹⁶ + 1949082795z³ y² x¹⁷ + 184365703z³ y x¹⁸ + 7391202z³ x¹⁹ + 683064z² y²⁰ + 16229412z² y¹⁹ x + 179090316z² y¹⁸ x² + 1226668788z² y¹⁷ x³ + 5872196907z² y¹⁶ x⁴ + 20949108255z² y¹⁵ x⁵ + 57928809309z² y¹⁴ x⁶ + 127368059034z² y¹³ x⁷ + 226421682549z² y¹² x⁸ + 328852313103z² y¹¹ x⁹ + 392354007216z² y¹⁰ x¹⁰ + 384932189046z² y⁹ x¹¹ + 309462418494z² y⁸ x¹² + 202136170716z² y⁷ x¹³ + 105692512158z² y⁶ x¹⁴ + 43211112153z² y⁵ x¹⁵ + 13318918194z² y⁴ x¹⁶ + 2922771246z² y³ x¹⁷ + 416043885z² y² x¹⁸ + 32877447z² y x¹⁹ + 1137108z² x²⁰ + 73608z y²¹ + 1920438z y²⁰ x + 22997691z y¹⁹ x² + 169821299z y¹⁸ x³ + 873530376z y¹⁷ x⁴ + 3345209124z y¹⁶ x⁵ + 9937822484z y¹⁵ x⁶ + 23530205745z y¹⁴ x⁷ + 45215279271z y¹³ x⁸ + 71346876950z y¹² x⁹ + 93075545721z y¹¹ x¹⁰ + 100626687732z y¹⁰ x¹¹ + 89991219620z y⁹ x¹² + 66139519362z y⁸ x¹³ + 39462842940z y⁷ x¹⁴ + 18738234743z y⁶ x¹⁵ + 6861256281z y⁵ x¹⁶ + 1841158926z y⁴ x¹⁷ + 331375220z y³ x¹⁸ + 33446001z y² x¹⁹ + 1137108z y x²⁰ + 2016y²² + 67284y²¹ x + 946386y²⁰ x² + 7881216y¹⁹ x³ + 44731008y¹⁸ x⁴ + 186741912y¹⁷ x⁵ + 601072962y¹⁶ x⁶ + 1538698518y¹⁵ x⁷ + 3199359072y¹⁴ x⁸ + 5479839330y¹³ x⁹ + 7799401764y¹² x¹⁰ + 9263965242y¹¹ x¹¹ + 9184308252y¹⁰ x¹² + 7569487044y⁹ x¹³ + 5141014410y⁸ x¹⁴ + 2835066066y⁷ x¹⁵ + 1240130304y⁶ x¹⁶ + 414702750y⁵ x¹⁷ + 99697038y⁴ x¹⁸ + 15350958y³ x¹⁹ + 1137108y² x²⁰) / (63 (y + 2x) (2y + 3x) (y² + y x + x²) (4y³ + 12y² x + 12y x² + 7x³) (z + y + 2x) (z + y + 3x) (z + 2y + 2x) (z + 2y + 3x) (2z + 3y + 3x) (z² + z y + z x + y² + 2y x + x²) (z² + 2z y + z x + y² + y x + x²) (3z³ + 9z² y + 9z² x + 9z y² + 18z y x + 9z x² + 3y³ + 9y² x + 9y x² + 7x³) (4z³ + 12z² y + 12z² x + 12z y² + 24z y x + 12z x² + 7y³ + 21y² x + 21y x² + 7x³))$$

Edit : For the hard case we use the same strategy on two variables this time ($x\leq y\le z$):

We have :

$$b(x,y,z)=\frac{x}{x+y}h\left(\frac{y}{x}\right)+\frac{y}{x+y}h\left(\frac{z}{y}\right)+\frac{\frac{13z^{4}}{x+y}}{8z^{3}+5x^{3}}-\frac{\left(x+y+z\right)}{x+y}+g\left(\frac{y+z}{x+y}\right)-h\left(\frac{y+z}{x+y}\right)\ge 0$$

Using Buffalo's way like Michael Rozenberg .

$$b(x,x+y,x+y+1)=(12584584192x²⁶ y² + 15798532096x²⁶ y + 12584584192x²⁶ + 231328988672x²⁵ y³ + 285675845632x²⁵ y² + 242958810624x²⁵ y + 80849636352x²⁵ + 2072986729216x²⁴ y⁴ + 2561101794816x²⁴ y³ + 2071033976832x²⁴ y² + 1254970977792x²⁴ y + 280449643776x²⁴ + 12061240907264x²³ y⁵ + 15255804251264x²³ y⁴ + 10625574437120x²³ y³ + 8537392776192x²³ y² + 4010158105344x²³ y + 664451078016x²³ + 51182491464064x²² y⁶ + 68035392184448x²² y⁵ + 37802179361472x²² y⁴ + 32320931462400x²² y³ + 25327574333568x²² y² + 9117338448768x²² y + 1180402133568x²² + 168657346263488x²¹ y⁷ + 241652708281344x²¹ y⁶ + 104283047058816x²¹ y⁵ + 65490015961248x²¹ y⁴ + 89073699394944x²¹ y³ + 56434851891648x²¹ y² + 15758650256832x²¹ y + 1650731664672x²¹ + 448546850939104x²⁰ y⁸ + 707591874538240x²⁰ y⁷ + 254116785725344x²⁰ y⁶ + 11459744865888x²⁰ y⁵ + 163487082564432x²⁰ y⁴ + 203393668597824x²⁰ y³ + 96951068044672x²⁰ y² + 21509499068896x²⁰ y + 1868659966384x²⁰ + 987748986709280x¹⁹ y⁹ + 1742055902427664x¹⁹ y⁸ + 618703132317744x¹⁹ y⁷ - 375162887424480x¹⁹ y⁶ - 19078174668384x¹⁹ y⁵ + 444159407170392x¹⁹ y⁴ + 362006378536208x¹⁹ y³ + 131344229466064x¹⁹ y² + 23719026176448x¹⁹ y + 1741396568616x¹⁹ + 1833187331695072x¹⁸ y¹⁰ + 3648068000649456x¹⁸ y⁹ + 1532110972781328x¹⁸ y⁸ - 1335334811814768x¹⁸ y⁷ - 1082846677998600x¹⁸ y⁶ + 463693139896536x¹⁸ y⁵ + 905100933152116x¹⁸ y⁴ + 500431394983152x¹⁸ y³ + 142713599779284x¹⁸ y² + 21426732852624x¹⁸ y + 1349400576504x¹⁸ + 2903349642153392x¹⁷ y¹¹ + 6545091866169152x¹⁷ y¹⁰ + 3561024481204640x¹⁷ y⁹ - 2629673344454264x¹⁷ y⁸ - 3648582187019564x¹⁷ y⁷ - 474046812091328x¹⁷ y⁶ + 1564142399854088x¹⁷ y⁵ + 1339877246007854x¹⁷ y⁴ + 544392172312490x¹⁷ y³ + 125839721338716x¹⁷ y² + 15986817970326x¹⁷ y + 873852157698x¹⁷ + 3958380338098336x¹⁶ y¹² + 10108497519673904x¹⁶ y¹¹ + 7221713929405896x¹⁶ y¹⁰ - 3119427484789240x¹⁶ y⁹ - 7377276657922682x¹⁶ y⁸ - 2989707012476520x¹⁶ y⁷ + 1833190413936286x¹⁶ y⁶ + 2709129511605578x¹⁶ y⁵ + 1490148552270996x¹⁶ y⁴ + 472333690860054x¹⁶ y³ + 90720121603248x¹⁶ y² + 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(3(2x + y)² (3x + y) (2x + 2y + 1) (3x + 3y + 1) (4x + 3y + 1) (6x + 4y + 1) (x² + x y + y²) (7x³ + 9x² y + 9x y² + 3y³) (x² + 2x y + x + y² + y + 1) (4x² + 6x y + 2x + 3y² + 3y + 1) (7x³ + 21x² y + 9x² + 21x y² + 18x y + 9x + 7y³ + 9y² + 9y + 3) (13x³ + 24x² y + 24x² + 24x y² + 48x y + 24x + 8y³ + 24y² + 24y + 8) (56x³ + 120x² y + 36x² + 96x y² + 72x y + 18x + 28y³ + 36y² + 18y + 3) (104x³ + 216x² y + 60x² + 168x y² + 120x y + 30x + 48y³ + 60y² + 30y + 5))$$

Second path for the hard case :

Let :

$$f\left(x\right)=\frac{1}{8+5x^{6}}$$

Then the function is convex on $(1.025,\infty)$ so using Jensen's inequality we have :

$$g\left(x,y,z\right)=\left(x^{2}+1\right)f\left(\frac{\left(\frac{z}{y}+xy\right)}{x^{2}+1}\right)+\frac{z^{8}}{8z^{6}+5x^{6}}-\frac{\left(x^{2}+y^{2}+z^{2}\right)}{13}$$

Next as the inequality is homogeneous we hav :

$$g(x,1,1+x+y)=(65x²⁰ + 480x¹⁹ y + 480x¹⁹ + 1680x¹⁸ y² + 3360x¹⁸ y + 2135x¹⁸ + 3040x¹⁷ y³ + 9120x¹⁷ y² + 12240x¹⁷ y + 6160x¹⁷ + 3400x¹⁶ y⁴ + 13600x¹⁶ y³ + 31080x¹⁶ y² + 34960x¹⁶ y + 15445x¹⁶ + 2480x¹⁵ y⁵ + 12400x¹⁵ y⁴ + 43840x¹⁵ y³ + 81920x¹⁵ y² + 78160x¹⁵ y + 30160x¹⁵ + 1160x¹⁴ y⁶ + 6960x¹⁴ y⁵ + 38400x¹⁴ y⁴ + 107200x¹⁴ y³ + 172200x¹⁴ y² + 148560x¹⁴ y + 49075x¹⁴ + 320x¹³ y⁷ + 2240x¹³ y⁶ + 21840x¹³ y⁵ + 86800x¹³ y⁴ + 212800x¹³ y³ + 309120x¹³ y² + 224000x¹³ y + 60800x¹³ + 40x¹² y⁸ + 320x¹² y⁷ + 8120x¹² y⁶ + 44240x¹² y⁵ + 162400x¹² y⁴ + 360640x¹² y³ + 448240x¹² y² + 289760x¹² y + 74275x¹² + 1920x¹¹ y⁷ + 13440x¹¹ y⁶ + 78960x¹¹ y⁵ + 260400x¹¹ y⁴ + 547200x¹¹ y³ + 707520x¹¹ y² + 471600x¹¹ y + 118320x¹¹ + 240x¹⁰ y⁸ + 1920x¹⁰ y⁷ + 24360x¹⁰ y⁶ + 119280x¹⁰ y⁵ + 546900x¹⁰ y⁴ + 1428240x¹⁰ y³ + 1801440x¹⁰ y² + 1044000x¹⁰ y + 221865x¹⁰ + 4800x⁹ y⁷ + 33600x⁹ y⁶ + 598860x⁹ y⁵ + 2658300x⁹ y⁴ + 4975800x⁹ y³ + 4563000x⁹ y² + 2011980x⁹ y + 336540x⁹ + 600x⁸ y⁸ + 4800x⁸ y⁷ + 680915x⁸ y⁶ + 4018290x⁸ y⁵ + 9688425x⁸ y⁴ + 12054700x⁸ y³ + 8105805x⁸ y² + 2766450x⁸ y + 368950x⁸ + 645130x⁷ y⁷ + 4515910x⁷ y⁶ + 13155990x⁷ y⁵ + 20620850x⁷ y⁴ + 18692390x⁷ y³ + 9721050x⁷ y² + 2649850x⁷ y + 285550x⁷ + 461075x⁶ y⁸ + 3688600x⁶ y⁷ + 12568715x⁶ y⁶ + 23771890x⁶ y⁵ + 27183875x⁶ y⁴ + 19110100x⁶ y³ + 7971005x⁶ y² + 1768450x⁶ y + 154435x⁶ + 242400x⁵ y⁹ + 2181600x⁵ y⁸ + 8514480x⁵ y⁷ + 18878160x⁵ y⁶ + 26110560x⁵ y⁵ + 23217600x⁵ y⁴ + 13137040x⁵ y³ + 4484400x⁵ y² + 814320x⁵ y + 57040x⁵ + 92670x⁴ y¹⁰ + 926700x⁴ y⁹ + 4076550x⁴ y⁸ + 10371600x⁴ y⁷ + 16847460x⁴ y⁶ + 18156600x⁴ y⁵ + 13029100x⁴ y⁴ + 6058000x⁴ y³ + 1705350x⁴ y² + 252460x⁴ y + 14230x⁴ + 25080x³ y¹¹ + 275880x³ y¹⁰ + 1350600x³ y⁹ + 3879000x³ y⁸ + 7241520x³ y⁷ + 9181200x³ y⁶ + 8002400x³ y⁵ + 4734400x³ y⁴ + 1826200x³ y³ + 424520x³ y² + 52120x³ y + 2920x³ + 4560x² y¹² + 54720x² y¹¹ + 295080x² y¹⁰ + 944400x² y⁹ + 1992840x² y⁸ + 2907840x² y⁷ + 2986760x² y⁶ + 2151600x² y⁵ + 1060800x² y⁴ + 341440x² y³ + 67680x² y² + 8640x² y + 920x² + 500x y¹³ + 6500x y¹² + 38280x y¹¹ + 135080x y¹⁰ + 317900x y⁹ + 524700x y⁸ + 620720x y⁷ + 527600x y⁶ + 317820x y⁵ + 132300x y⁴ + 37800x y³ + 8520x y² + 2020x y + 340x + 25y¹⁴ + 350y¹³ + 2235y¹² + 8620y¹¹ + 22385y¹⁰ + 41250y⁹ + 55315y⁸ + 54440y⁷ + 39275y⁶ + 20850y⁵ + 8625y⁴ + 3340y³ + 1355y² + 430y + 65)\ge 0$$

As $x,y,z>0$ and I only set the numerator .

There is a little typo it's at the end $g(x,1,x+y)=(4286x^{20}+22992x^{19}y+70824x^{18}y^{2}+28394x^{18}+125248x^{17}y^{3}+147840x^{17}y+140080x^{16}y^{4}+449664x^{16}y^{2}+80358x^{16}+102176x^{15}y^{5}+784448x^{15}y^{3}+404208x^{15}y+47792x^{14}y^{6}+865200x^{14}y^{4}+1210680x^{14}y^{2}+186050x^{14}+13184x^{13}y^{7}+622944x^{13}y^{5}+2076480x^{13}y^{3}+1297440x^{13}y+1648x^{12}y^{8}+288400x^{12}y^{6}+2249520x^{12}y^{4}+5345040x^{12}y^{2}+113690x^{12}+79104x^{11}y^{7}+1591968x^{11}y^{5}+14115360x^{11}y^{3}+514800x^{11}y+9888x^{10}y^{8}+726768x^{10}y^{6}+26769360x^{10}y^{4}+1456680x^{10}y^{2}+65886x^{10}+197760x^{9}y^{7}+38249880x^{9}y^{5}+2303520x^{9}y^{3}+286272x^{9}y+24720x^{8}y^{8}+41902280x^{8}y^{6}+2217300x^{8}y^{4}+795744x^{8}y^{2}+20354x^{8}+35305570x^{7}y^{7}+1296660x^{7}y^{5}+1245888x^{7}y^{3}+82320x^{7}y+22784135x^{6}y^{8}+390255x^{6}y^{6}+1211280x^{6}y^{4}+219144x^{6}y^{2}+2678x^{6}+11144640x^{5}y^{9}-18960x^{5}y^{7}+761376x^{5}y^{5}+323008x^{5}y^{3}+9888x^{5}y+4054380x^{4}y^{10}-69480x^{4}y^{8}+311472x^{4}y^{6}+288400x^{4}y^{4}+24720x^{4}y^{2}+1062240x^{3}y^{11}-28800x^{3}y^{9}+79104x^{3}y^{7}+161504x^{3}y^{5}+32960x^{3}y^{3}+189480x^{2}y^{12}-5880x^{2}y^{10}+9888x^{2}y^{8}+57680x^{2}y^{6}+24720x^{2}y^{4}+20600xy^{13}-720xy^{11}+13184xy^{7}+9888xy^{5}+1030y^{14}-40y^{12}+1648y^{8}+1648y^{6})\ge0$$

$\endgroup$
1

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