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

[lfdahl]

Let $a, b, c$ be non-negative real numbers satisfying $a + b + c = 5$.

Find the maximal value of $a^4b+b^4c+c^4a$.

 

[lfdahl]

Suggested solution:

Spoiler

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.

 

[kaliprasad]

Suggested solution:

Spoiler

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.

Spoiler

Three triples are missing solution set is $(4,1,0),(4,0,1),(0,1,4), (0,4,1),(1,4,0),(1,0,4)$

 

[Opalg]

Spoiler

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]

 

[kaliprasad]

[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