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
 A:    →     →   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.
A: 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) $2.6<c\le3$: 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.
