How to prove this inequality $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$?

Question:

If $a,b,c$ are nonnegative real numbers such that $a+b+c=3,$ then

$$(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$$

My try: I found the equality holds only if $(a,b,c)=(2,0,1)$ or all of its permutations.

But I can't prove this inequality it. I would appreciate very much a proof.

This problem comes from:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=562119

• Do you mean "I found that equality holds only if $(a,b,c)$ is $(2,0,1)$, $(1,2,0)$, or $(0,1,2)$"?. – John Bentin Nov 13 '13 at 11:05
• If you can prove one of them is zero, then you can just study $f(2+x,0,1-x)$ and see that its maximum is $64$. – Beni Bogosel Nov 13 '13 at 12:20
• You can edit your question so that it says what you mean. – John Bentin Nov 13 '13 at 12:26
• @math110: I've taken the liberty of editing it for you. Where did you get this question from, if you don't mind me asking. – Nick Nov 13 '13 at 13:05 → → The three dimensional problem can be simplified to a two dimensional problem by introducing (again and again) suitable triangle coordinates: $$\left[ \begin{array}{c} a \\ b \\ c \end{array} \right] = \left[ \begin{array}{c} 3 \\ 0 \\ 0 \end{array} \right] + \left[ \begin{array}{c} -3 \\ 3 \\ 0 \end{array} \right] x + \left[ \begin{array}{c} -3 \\ 0 \\ 3 \end{array} \right] y$$ Then the equation $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$ does not so much "simplify", but anyway becomes an equation in two variables (2-D). And the equation $\;a + b + c = 3\;$ corresponds with a normed 2-D triangle, with vertices $(0,0),(1,0),(0,1)$ . The insides of both can easily be visualized, as has been done in the above picture in the middle:
$\color{red}{red}$ for $\;a + b + c = 3\;$ and $\color{green}{green}$ for $\;(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64\;$ .
The transformed inequality is: $$\left[ 3^2\left( 1-x-y \right)^{2} + 3^5\,x{y}^{4} \right] \left[ 3^2\,{x}^{2} + 3^5\,y \left( 1-x-y \right)^{4} \right] \left[ 3^2\,{y}^{2} + 3^5\, \left( 1-x-y \right){x}^{4} \right] \le 64$$

It is seen in the same picture that the edge $\;y=1-x\;$ of the triangle maybe is tangent to the curve $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) = 64$ . Indeed, if we substitute $y=1-x$ into (the transformation of) that equation and simplify, then we get: $$3^9\, x^3 (1-x)^6 - 64 = 0$$ The same sort of equation is found with the substitutions $\,x=0\,$ or $\,y=0\,$, for the other two edges. And, as has been found by others, there is only one solution of that equation, within the specified range, namely $x=1/3$, corresponding with $y=2/3$ and hence $(a,b,c) = (0,1,2)$ . And of course any cyclic permutation of this, due to symmetry. The rest of the (red) triangle $\;a + b + c = 3\;$ is well within the (green) area of $\;(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$ . Which can be shown by plotting the triangle first: then it becomes absorbed by the green area (see picture on the right).

Analytically, the proof is completed by considering the function $\;f(x) = 3^9\, x^3 (1-x)^6$ .
Its extreme values are found for $\;f'(x) = x^2(1-x)^5(3-9x)=0$ , giving $\;x = \{0,1,1/3\}$ , with the maximum $\;f(1/3)=64$ .

The picture on the right shows the the inequality as observed in the plane $\;a + b + c = 3\;$ of the triangle in 3-D (picture on the left). Mind the symmetries.

• Can you use other colors? Since many people have protanopia or protanomaly (like me). – ziyuang Jan 8 '14 at 11:24
• Next time, I promise. Any suggestion? I've used red and green now. – Han de Bruijn Jan 8 '14 at 14:45
• Maybe different patterns of lines, if that's convenient for you. – ziyuang Jan 8 '14 at 23:05

Without loss of generality, assume $$a$$ is smallest of $$a, b, c$$. Also, let $$f(a,b,c)=(a^2+bc^4)(b^2+ca^4)(c^2+ab^4)$$and firstly, if $$a\le c\le b$$, then$$f(a,b,c)-f(a,c,b)=(b^3-a^3) (c^3 - a^3) (b ^3- c^3) (a b c - 1)<0$$therefore we can assume $$a\le b\le c$$. Now, we will prove $$f(0,b,a+c)\ge f(a,b,c)$$which is, after full expansion,$$a (-a^6 b^4 c + a^5 b^3 - a^5 c^3 - a^4 b^5 c^5 - a^3 b c^7 - a^2 b^6 + 20 a^2 b^3 c^3 \\+ 15 a b^3 c^4 - a b^2 c^2 - b^7 c^4 + 6 b^3 c^5)+ 6 a^5 b^3 c + 15 a^4 b^3 c^2\ge0$$ and it is obvious that $$-a^6 b^4 c + a^5 b^3\ge0$$, $$- a^5 c^3 - a^4 b^5 c^5 - a^2 b^6 + 20 a^2 b^3 c^3\ge0$$ and $$6 a^5 b^3 c + 15 a^4 b^3 c^2\ge0$$. Therefore it is enough to show$$b c^2 (-a^3 c^5 + 15 a b^2 c^2 - a b - b^6 c^2 + 6 b^2 c^3)\ge0$$and from $$-a^3c^5\ge-ab^2c^5$$, $$-ab\ge-abc^3\ge-b^2c^3$$ and $$-b^6c^2>-4b^3c^2\ge-4b^2c^3$$, it is left to show$$b^2c^2(-a c^3 + 15 a + c)\ge0$$and we can divide it with three cases.

Case 1) $$c^3\le15$$: $$-a c^3 + 15 a + c>a(15-c^3)\ge0$$.

Case 2) $$15^{1/3}< c\le2.6$$: Firstly, $$c>2.4$$. The equation is decreasing with respect to $$a$$, therefore we need to show only for maximal value of $$a$$. If $$a>0.3$$, then $$a+b+c\ge2a+c>0.6+2.4=3$$, so maximal value is $$0.3$$. Also, $$c^3\le2.6^3<20$$. Therefore, $$-a c^3 + 15 a + c\ge0.3(15-c^3)+c\ge-1.5+2>0$$

Case 3) $$2.6: Similarly, it is enough to show for maximal value of $$a$$ which is $$0.2$$. Therefore, $$-a c^3 + 15 a + c\ge0.2(15-27)+2.6>0$$

Therefore, we can assume that $$0=a\le b\le c$$. Now $$f(a,b,c)=b^3c^6\le2^6\left(\frac{3\times b+6\times0.5c}{9}\right)^9=64$$ and it is proved.