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let $a,b,c>0$,and such that $a+b+c=3$,prove that $$(abc)^4+abc(a^3c^2+b^3a^2+c^3b^2)\le 4$$

I first consider $$abc\le\left(\dfrac{a+b+c}{3}\right)^3=1$$ so it suffices to show that $$a^3c^2+b^3a^2+c^3b^2\le 3$$ But I find this not true.

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have you tried using Lagrnge multipliers to find the minimum of the target function? See this… – yohBS Apr 15 '13 at 6:36
I think this Lagrnge multipliers is very ugly. – math110 Apr 15 '13 at 6:45
I agree. But it works. – yohBS Apr 15 '13 at 6:51
this symterm are very hard! – math110 Apr 15 '13 at 7:16
I think your inequality is equal to this inequality :D – pxchg1200 Apr 18 '13 at 4:33
up vote 2 down vote accepted

I copy a special method to prove $abc+abc(a^3b^2+b^2c^3+c^3a^2) \leq 4$ which is equal to current one. $abc+abc(a^3b^2+b^2c^3+c^3a^2) \leq 4$$ \Leftrightarrow4(a+b+c)^8-6561abc(a^3c^2+b^3a^2+c^3b^2)-27abc(a+b+c)^5\geq0 $

$ a=\min\{a,b,c\} , b=a+u ,c=a+v $

$ 4(a+b+c)^8-6561abc(a^3c^2+b^3a^2+c^3b^2)-27abc(a+b+c)^5= $$ =4374(u^2-uv+v^2)a^6+243(49u^3+21u^2v-60uv^2+49v^3)a^5+ 81(164u^4+242u^3v-168u^2v^2-163uv^3+164v^4)a^4+ 27(208u^5+707u^4v+352u^3v^2-1106u^2v^3+221uv^4+208v^5)a^3+9(109u^6+609u^5v+1455u^4v^2-1006u^3v^3-732u^2v^4+609uv^5+109v^6)a^2+ +3(32u^7+215u^6v+627u^5v^2+1030u^4v^3-1157u^3v^4+627u^2v^5+215uv^6+32v^7)a+4(u+v)^8\geq0 $


let $K_{i}$means $K_{i}a^i, i=1 to 6$, If $K_{i} \ge 0$, then <1> will be true.

$K_6=u^2-uv+v^2 \ge 0$ is trivial . $K_5=49u^3+21u^2v-60uv^2+49v^3$, $21u^2v+49v^3 \ge 2\sqrt{21*49}uv^2>2*32uv^2>60uv^2$ $\to K_5 \ge 0$ $K_4=164u^4+242u^3v-168u^2v^2-163uv^3+164v^4=164u^4+78u^3v-4u^2v^2+uv^3+164v(u^3-u^2v-uv^2+v^3)$, $(u-v)(u^2-v^2) \ge 0 \to u^3-u^2v-uv^2+v^3 \ge 0, 78u^3v+uv^3 \ge 2\sqrt{78*1}u^2v^2 > 4u^2v^2 \to K_4 \ge 0$ $K_3=208u^5+707u^4v+352u^3v^2-1106u^2v^3+221uv^4+208v^5$,$352u^3v^2+221uv^4 \ge 2\sqrt{352*221}u^2v^3>557u^2v^3,707u^4v+208v^5 \ge 2\sqrt{707*208}u^2v^3>766u^2v^3,557+766=1323>1106 \to K_3 \ge 0$ $K_2=109u^6+609u^5v+1455u^4v^2-1006u^3v^3-732u^2v^4+609uv^5+109v^6$,$609u^5v+609uv^5 \ge 1218u^3v^3>1006u^3v^3,1455u^4v^2+109v^6 \ge 2\sqrt{1455*109}u^2v^4>796u^2v^4>732u^2v^4 \to K_2 \ge 0$ $K_1=32u^7+215u^6v+627u^5v^2+1030u^4v^3-1157u^3v^4+627u^2v^5+215uv^6+32v^7,1030u^4v^3+627u^2v^5 \ge 2\sqrt{1030*627}u^3v^4 >1607u^3v^4 >1157u^3v^4 \to K_1 \ge 0 $


the original proof is here

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It is possible to solve this by $pqr$ method. (That is, write everything in terms of $p = a+b+c$, $q=ab+bc+ca$, $r = abc$) together with $uvw$ method. This is an ugly method, but it works. The approach is sketched here.

  • Although the inequality is not symmetric, one can write $$\begin{array} &&2(a^3c^2 + b^3a^2 + c^3b^2) \\ &= a^3(b^2+c^2) + b^3(a^2+c^2) + c^3(a^2+b^2) + (a^3c^2+b^3a^2 + c^3b^2 - a^3b^2 - b^3c^2 - c^3a^2) \\ &= a^3(b^2+c^2) + b^3(a^2+c^2) + c^3(a^2+b^2) + (b-a)(c-a)(c-b)(ab+bc+ca) \end{array}$$

This allows one to rewrite the inequality as $$2(abc)^4 + abc(\sum_{sym} a^3b^2) + abc(ab+bc+ca)\sqrt{(a-b)^2(b-c)^2(c-a)^2} \leq 8$$

  • I used Sage to express the expression in terms of $p,q,r$, which hopefully someone can verify:

$$2r^4 + r(pq^2 - 2p^2r - qr) + qr \sqrt{\frac{4(p^2-3q)^3 - (2p^3-9pq+27r)^2}{27}} \leq 8$$

The main problem here, would be to efficiently remove the square root. For this, we use a lemma by Vasile Cirtoaje and Vo Quoc Bao Can, lemma 2.1 in here. In the lemma we take $$\alpha = \frac{1}{\sqrt{27}}, a = 2(9-3q)^{3/2}, b = 54-27q+27r, \beta = \frac{14+q}{27q}$$ (The choice of $\alpha,a,b$ comes from the shape of the square root, while the choice of $\beta$ comes from cancelling the "linear term" of $r$ in $pq-2p^2r-qr = 3q - (18+q)r$; remember that $p = 3$ in this question) Therefore LHS is bounded above by $$2r^4 + 3rq^2 + (27q\frac{14+q}{27q} - (18+q))r^2 + r(q\frac{14+q}{27q} (54-27q) + 2(9-3q)^{3/2} \sqrt{\frac{q^2}{27} + \left(\frac{14+q}{27}\right)^2}) ...(*)$$ and it suffices to show that $(*) \leq 8$.

  • For this, we use the uvw method. Fix $r$. By the help of a computer, one can check that $(*)$ is convex in $q$. Thus to maximize $(*)$, $q$ should take maximum/minimum of its allowable range, which would force two of $a,b,c$ to be equal according to $uvw$ method.

  • Assume WLOG that $a=c$. Then $2a+b = 3$, and $a^2b = r$. Solving these gives $$b = 3-2a \, \text{ and } \, r = a^2(3-2a) \, \text{ and } \, q = -3a^2+6a$$ The only constraint we have on $a$ are $a \in [0,3/2]$ (coming from $2a+b=3$ and $a,b \ge 0$), and $a^2(3-2a) = r \leq 1$ (since $p = 3$). The latter inequality is equivalent to $a \ge -1/2$, which means that the only constraint we have on $a$ is $a \in [0,3/2]$.

  • Therefore we are reduced to showing that for $a \in [0,3/2]$, the substituted expression $(**) \leq 8$, where $(**)$ comes from substituting $q,r$ into $(*)$ in terms of $a$. I find that $(**)$ is the following, and please verify this if (quite unlikely) you are reading up to this line. $$2 (a^2 (3-2 a))^4+3a^2(3-2a)(-3a^2+6a)^2 - 4(a^2(3-2a))^2 + a^2(3-2a)(14 - 3a^2 + 6a)(2 - (-3a^2+6a))+2a^2 (3-2 a) (9-3 (-3 a^2+6 a))^{3/2} \sqrt{\frac{(-3a^2+6a)^2}{27}+\frac{(14-3 a^2+6 a)^2}{27^2}}..(**)$$ This is a single variable calculus problem that is ugly but should be tractable. I trust WolframAlpha to do the job for me, which does give the maximum is 8.

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this is good job! – math110 Apr 19 '13 at 4:59
a good method. if we can conclude the maximum will be only get when a=c,then things will be simpler and much more, we have $abc+abc(a^3c^2+b^3a^2+c^3b^2)\le 4$ also OK. but if I use the link "job" , when I reduce the "4" to "1", the wolframAlpha will give a wrong number(max>8 but a=.0566), and if I input the $abc+abc(a^3c^2+b^3a^2+c^3b^2)$ directly, I will get right answer that max is 4. I don't know why. – chenbai Apr 21 '13 at 4:17
@chenbai, I used an estimate in the middle (citing Vasile Cirtoaje and Vo Quoc Bao Can), so the last inequality implies but is not equivalent to the original inequality. When you replace 4 by 1, $r \ge r^4$, so things go up a bit. If you look at the link "job", you can see that the peak at 1 is not much larger than a local peak at ~0.5, so it's conceivable that as you change 4 to 1, the gain you have there is large enough to create a new and bigger maximum. – user27126 Apr 21 '13 at 6:17
Thanks a lot.I learn a lot from your post. – chenbai Apr 21 '13 at 12:43
@Sanchez do you mind sharing the code that you used in Sage? – Vincent Tjeng Apr 23 '13 at 7:53

EDit: 5th version: After read "uvw method", I found the trick is:$abc(a^3c^2+b^3a^2+c^3b^2)\leq 3$, but I have assumption that $a \ge b \ge c$,and all my explore is based on it but it is wrong. thanks Sanchez points out my mistake which force me to recheck everything and I get a real result $a=1.3175,b=0.5856,c=1.0969$ that cause $abc(a^3c^2+b^3a^2+c^3b^2)=3.00667>3$ ,Indeed, the $abc(a^3c^2+b^3a^2+c^3b^2)\leq 3$ will be true when $(a-b)(b-c)(c-a) \leq 0 $, if $(a-b)(b-c)(c-a) \ge 0 $, then $abc(a^3b^2+b^3c^2+c^3a^2)\leq 3$ will be true.that is a real trick i found. To proof it ,just use Sanchez's work:

$2abc(a^3c^2+b^3a^2+c^3b^2)=abc(\sum_{sym} a^3b^2) + abc(ab+bc+ca)(a-b)(b-c)(c-a) \leq 6$, I will proof $abc(\sum_{sym} a^3b^2) \leq 6$ next. then the result is clear. if we swap $b,c$ , we have another result at once.

Here I still use Lagrange multiplier to proof $abc(a^3c^2+b^3a^2+c^3b^2+a^3b^2+a^2c^3+b^3c^2) \leq 6$.

let $f=abc(a^3c^2+b^3a^2+c^3b^2+a^3b^2+a^2c^3+b^3c^2)- 6$, $g=a+b+c-3$, $F=f-\lambda g ,p=a^3c^2+b^3a^2+c^3b^2+a^3b^2+a^2c^3+b^3c^2,$

$\dfrac{\partial F}{\partial a}=bcp+abc[3a^2(b^2+c^2)+2a(b^3+c^3)]= \lambda $......<1>

$\dfrac{\partial F}{\partial b}=acp+abc[3b^2(a^2+c^2)+2b(a^3+c^3)]= \lambda $......<2>

$\dfrac{\partial F}{\partial c}=abp+abc[3c^2(a^2+b^2)+2c(a^3+b^3)]= \lambda $......<3>

$\dfrac{\partial F}{\partial \lambda}=a+b+c-3=0$......<4>

<1> - <2> : $pc(b-a)+abc[3c^2(a^2-b^2)+2ab(b^2-a^2)+2c^3(a-b)]=0$

that is:


so we have $c=0$ or $a=b$ or


we look at <6> carefully,

$c^3(a^2+b^2) \ge 2c^3ab $, $3a^2b^2(a+b) \ge 3abc^2(a+b)$,so RHS $> $ LHS, so we can ignore<6>.

the $c=0$ is the max value of $a^3c^2+b^3a^2+c^3b^2+a^3b^2+a^2c^3+b^3c^2$, that is why $a^3c^2+b^3a^2+c^3b^2<3$ is not true. we can ignore here as it will lead f=-6. so we have a=b,

In case $a=b $, let <1>-<3>, we have:


we have $a=c$, or $[p+bc(2ac(a+c)-3b^2(a+c)-2b^3)]=0$ which is:


put $a=b, 2b+c=3$, we simplify<8> to:

$10 b^3-84 b^2+153 b-81 = 0 $......<9>,

$q(b)=10 b^3-84 b^2+153 b-81,q'(b)=30b^2-168b+153=0,$......<10>

get two roots: $b_1 = \dfrac{1}{10}(28 - \sqrt{274}), b_2 = \dfrac{1}{10}(28 + \sqrt{274}>4)$,

put $b=0$ and $b=2$ into<10>,$q'(0)>0,q'(2)<0,$ so $b_1$ is max point between $(0,2)$, put $b_1$ into $q(b_1) =-.9<0$ ,so there is no solution for $0<b \leq 1.5$ so the only solution is $a=c$ for <7>.

last step, we verify , when $a=b=c=1,f=0,$ we already have $-6$ before,so $f_{max}=0$. QED.

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Thank you,But I think this :we know with linear caculation, we must have: $P_{11}u+P_{12}v+P_{13}w=0$ $P_{21}u+P_{22}v+P_{23}w=0$ $P_{31}u+P_{32}v+P_{33}w=0$ from <8>,<9>,<10> we know, $\begin{vmatrix} P_{11} & P_{12} & P_{13} \\ P_{21} & P_{22} & P_{23}\\ P_{31} & P_{32} & P_{33} \end{vmatrix}\not=0$ ,so we must have: $u=0,v=0,w=0,$ ie $\lambda=0 $ or $ a=b=c$ have some problem,why? can you explain ? – math110 Apr 16 '13 at 10:46
Please elaborate on what $P_{ij}$ are and why the determinant of that $3 \times 3$ matrix is supposed to be $0$. By themselves, equations <8>, <9>, <10> do not necessarily imply that $u=v=w=0$, unless you somehow used the definition of $X, Y, Z, u, v, w$ without elaborating, in which case you should provide more justification. – Ivan Loh Apr 16 '13 at 10:53
This argument is flawed. Consider the paragraph starting with "now we can deduct X,Y,Z, for example, $\ldots$" I shall actually do the algebraic manipulation. $<8> -<9>\times 4$: $$-13Y-11Z+u-4v=0 \ldots <11>$$ $<8> \times 3 - <10> \times 4$: $$5Y-13Z+3u-4w=0 \ldots <12>$$ $<9> \times 3 - <10>$: $$11Y+5Z+3v-w=0 \ldots <13>$$ $<11> \times 5+<12> \times 13$: $$-224Z+44u-20v-52w \ldots <14>$$ $<11> \times 11+<13> \times 13$: $$-56Z+11u-5v-13w \ldots <15>$$ $<12> \times 11-<13> \times 5$: $$-168Z+33u-15v-39w=0 \ldots <16>$$ Now $<14>, <15>, <16>$ are scalar multiples of each other. – Ivan Loh Apr 16 '13 at 13:53
If you now try to remove $Z$ by considering a linear combination of the last 3 equations above, we will just get $0u+0v+0w=0$, so your $P_{ij}$ are all $0$, and nothing can be concluded. – Ivan Loh Apr 16 '13 at 13:57
It's wrong. When the first inequality is violated, you can find SOME a,b,c, not ALL a,b,c, such that $abc(a^3c^2 + b^3a^2 + c^3b^2) > 3$. There's no guarantee that the triple $(a,c,b)$ would still violate the inequality - you can't just say "x+z+y = 3" so this is true. – user27126 Apr 23 '13 at 7:29

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