MHB Find the maximal value of a^4b+b^4c+c^4a

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The discussion focuses on maximizing the expression a^4b + b^4c + c^4a under the constraint a + b + c = 5, with a, b, and c being non-negative real numbers. Participants note that while there is cyclic symmetry in the variables, it does not imply complete symmetry, leading to different values for various permutations of a, b, and c. A suggested solution was initially deemed incorrect due to this misunderstanding of symmetry. The conversation highlights the importance of recognizing the specific characteristics of the function when attempting to find its maximum value. The goal remains to determine the maximal value of the expression given the constraints.
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Let $a, b, c$ be non-negative real numbers satisfying $a + b + c = 5$.

Find the maximal value of $a^4b+b^4c+c^4a$.
 
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Suggested solution:
The maximal value is $256$ and is attained at $(4,1,0), (0,4,1)$ or $(1,0,4)$.

Define $f(x,y,z) = x^4y+y^4z+z^4x$. Let $a \geq b$ and $a\geq c$.

Let us prove, that $f(a+c/2,b+c/2,0) \geq f(a,b,c)$. Indeed,

\[f(a+c/2,b+c/2,0) = (a+c/2)^4(b+c/2) \geq (a^4+2a^3c)(b+c/2) \geq a^4b+2a^3bc+a^3c^2 \geq a^4b+b^4c+c^4a = f(a,b,c)\]

Now, we maximize $f(a,b,0)$, when $a+b = 5$ by using the AM-GM inequality:

$5 = a+b = (a/4+a/4+a/4+a/4 + b) \geq 5\sqrt[5]{a^4b4^{-4}}$.

Therefore, $a^4b \geq 4^4$. Equality holds at $a = 4, b=1$. Similarly we obtain other maximum triples $(0,4,1)$ and $(1,0,4)$ when maximum of $a,b$ and $c$ is $b$ and $c$. Done.
 
lfdahl said:
Suggested solution:
The maximal value is $256$ and is attained at $(4,1,0), (0,4,1)$ or $(1,0,4)$.

Define $f(x,y,z) = x^4y+y^4z+z^4x$. Let $a \geq b$ and $a\geq c$.

Let us prove, that $f(a+c/2,b+c/2,0) \geq f(a,b,c)$. Indeed,

\[f(a+c/2,b+c/2,0) = (a+c/2)^4(b+c/2) \geq (a^4+2a^3c)(b+c/2) \geq a^4b+2a^3bc+a^3c^2 \geq a^4b+b^4c+c^4a = f(a,b,c)\]

Now, we maximize $f(a,b,0)$, when $a+b = 5$ by using the AM-GM inequality:

$5 = a+b = (a/4+a/4+a/4+a/4 + b) \geq 5\sqrt[5]{a^4b4^{-4}}$.

Therefore, $a^4b \geq 4^4$. Equality holds at $a = 4, b=1$. Similarly we obtain other maximum triples $(0,4,1)$ and $(1,0,4)$ when maximum of $a,b$ and $c$ is $b$ and $c$. Done.

Three triples are missing solution set is $(4,1,0),(4,0,1),(0,1,4), (0,4,1),(1,4,0),(1,0,4)$
 
kaliprasad said:
Three triples are missing solution set is $(4,1,0),(4,0,1),(0,1,4), (0,4,1),(1,4,0),(1,0,4)$
[sp]Not true: $f(4,0,1) = f(1,4,0) = f(0,1,4) = 4$. There is cyclic symmetry but not complete symmetry in the variables.

[/sp]
 
Opalg said:
[sp]Not true: $f(4,0,1) = f(1,4,0) = f(0,1,4) = 4$. There is cyclic symmetry but not complete symmetry in the variables.

[/sp]

Oops my mistake. I did not realize the output is non symmetric
 
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