# Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$

$$x,y,z >0$$, 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".

• 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. – HN_NH May 7 '16 at 16:37
• 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.) – Ege Erdil May 7 '16 at 23:01
• 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... – Nicolas May 22 '16 at 9:13
• Any further ideas for changes of the tags here must be first proposed to me. – quid Oct 13 '18 at 23:22
• @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?) – Martin R Dec 29 '19 at 13:23

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 $$69589^2-4\cdot131015\cdot9633<0.$$

• 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. – miracle173 Jul 3 '16 at 12:16
• there is an error in my edit. One cannot assume $z\le y$. I try do undo the change. – miracle173 Jul 3 '16 at 13:06
• @MichaelRozenberg: can you elaborate how you get the the polynomial in t and the estimates? – miracle173 Jul 3 '16 at 13:22
• @miracle173 for example, $156(u^2-uv+v^2)x^8\geq156uvx^8=156u^5v^5t^8$. – Michael Rozenberg Jul 3 '16 at 13:31
• Here is a link to an answer that contains a Maxima script with some details to this answer. – miracle173 Jul 3 '16 at 18:30

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 $\,0 \le f(x,y,z) < 0.00002$ .

• Interesting technique. Thank you very much for trying my inequality. One up vote !!! – HN_NH May 24 '16 at 6:05

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.

• So directly using CS is too weak. Any other stronger inequalities that we can try? – Colescu May 22 '16 at 5:56
• 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. – user126154 Jun 23 '16 at 9:03
• 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 – Mr Pie Feb 2 '19 at 13:58

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 wand 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)
(%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 wand 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 wand 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.

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...

• Thank you very much for using Lagrange Multiplier ! One up vote from me :) – HN_NH May 24 '16 at 6:06
• @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. – Nicolas May 24 '16 at 9:07

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!

• +1 ; about time you answered it by Holder with polynomial integers :> – user552223 Jun 7 '20 at 10:31
• You did it by your algorithm, didn't you ? – user552223 Jun 7 '20 at 10:33
• @Giang Nguyễn Đặng Thanh I just played with coefficients and it turned out. – Michael Rozenberg Jun 7 '20 at 10:48
• @Giang Nguyễn Đặng Thanh I tried, but without success. I posted a proof of this inequality here: math.stackexchange.com/questions/3589716 – Michael Rozenberg Jun 7 '20 at 12:38
• (+1) Nice C-S, now the form is better. – River Li Jun 7 '20 at 14:19

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.

• 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? – Michael Rozenberg Jan 5 '20 at 3:21

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). $$\sum_{k=1}^{n}{\frac{x_{k}^{2}}{a_{k}}} \ge \frac{\left(\sum_{k=1}^{n}{x_{k}}\right)^{2}}{\sum_{k=1}^{n}{a_{k}}}$$

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 $\beta=5$.

• 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. – NivPai Aug 22 '18 at 16:22
• Yes, it is also known by the name Bergstrom’s inequality. There is also a slightly tighter generalization as well. – Rethna Pulikkoonattu Aug 22 '18 at 16:35
• 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 $2X^2Y^2-X^3Y-XY^3=-XY(X-Y)^2.$ – Jose Brox Aug 22 '18 at 16:37
• Thanks @Jose Brox. Yes, then there is a hole in my argument. We still have it open then:-) – NivPai Aug 22 '18 at 17:04
• @ 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. – NivPai Aug 22 '18 at 17:20

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 $0\le a,b\le 1$

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.....

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

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

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

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

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}

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

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

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).

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

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),

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

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

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).

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

PROVED.

• Dear @Yuri Negometyanov I don't see a connection between your paper and starting inequality. I am sorry. – Michael Rozenberg Jun 22 '17 at 2:40
• @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 $a$. – River Li Jan 3 '20 at 3:55
• @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 $k=^{133}\!/_{81}.$ – Yuri Negometyanov Jan 25 '20 at 15:51
• @YuriNegometyanov Please explain the 3rd group. – River Li Jan 25 '20 at 16:19
• @YuriNegometyanov Nice. I got it. – River Li Jan 26 '20 at 14:23

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, $$x^2 + y^2 + z^2 \ge -3 + 3 + (x + y + z)$$ $$x^2 + y^2 + z^2 \ge x + y + z \ge 3$$

• 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 $z=\frac{1}{2}.$ – Michael Rozenberg Aug 26 '18 at 1:28
• 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 $xyz=1$. – Vee Hua Zhi Aug 26 '18 at 2:04
• 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 $y=z=0.5$. – Michael Rozenberg Aug 26 '18 at 6:35
• OK I get it @MichaelRozenberg – Vee Hua Zhi Aug 28 '18 at 12:47

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 !

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 $$k\frac{1}{5a^3+8}-l\frac{b}{8b^3+5}\leq (k-l)\frac{a}{8a^3+5b^3}\Leftrightarrow$$ $$\frac{k}{5k+13}-\frac{lb}{8l+13}\leq (k-l)\frac{a}{8k+5l-13}\Leftrightarrow$$ $$\frac{k}{k-l}\frac{1}{x_1}-\frac{l}{k-l}\frac{b}{y_1}\leq \frac{a}{x_1+y_1-13}$$ Or equivalent (we set $$x_1=5k+13$$, $$y_1=8l+13$$) if: $$f_0(x_1,y_1):=ky_1^2-blx_1^2+x_1y_1(-bl+k-ak+al)-13ky_1+13lb x_1\leq 0.$$ 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 $$x_0=\frac{13k(k-ak+(a+b)l)}{(a-1)^2k^2+2(b+a(1-a+b))kl+(a-b)^2l^2}$$ $$y_0=\frac{13bl(k+ak+(-a+b)l)}{(a-1)^2k^2+2(b+a(1-a+b))kl+(a-b)^2l^2}$$ are the points such that $$\partial_xf_0(x_0,y_0)=\partial_yf_0(x_0,y_0)=0$$ and $$f_0(x_0,y_0)=\frac{169abkl(l-k)}{(a-1)^2k^2+2(b+a(1-a+b))kl+(a-b)^2l^2}\leq 0$$ and $$(a-1)^2k^2+2(b+a(1-a+b))kl+(a-b)^2l^2>0$$, $$a>1$$, $$a\geq b\geq 1$$. 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

• You consider the case $x\ge y\ge z$, right? How about the left case? – River Li Dec 25 '20 at 1:07
• @RiverLi. Thanks for the note. Can you check it now please? – Nikos Bagis Dec 26 '20 at 23:35
• 1) It should be $\Pi(x,y,z)=\lambda^{-1}\Pi(\lambda x,\lambda y,\lambda z)=z\Pi\left(\frac{x}{z},\frac{y}{z},1\right)$. – River Li Dec 27 '20 at 4:26
• 2) In (2), is should be $\Pi(x,y,z)=z\Pi(a,b,1)= \frac{5}{13}z (\frac{1-a^3}{8+5a^3}+\frac{b(-1+b^3)}{5+8b^3}+\frac{a^4-ab^3}{8a^3+5b^3})$ – River Li Dec 27 '20 at 4:26
• 3) In (3), $f(1,a)(1-a^3)\geq f(b,1)(1-b^3)$ is not true (check $a = 2$, $b = 1$). – River Li Dec 27 '20 at 4:40