The discussion revolves around finding the maximal value of the expression \(a^4b + b^4c + c^4a\) under the constraint that \(a, b, c\) are non-negative real numbers summing to 5. The focus includes mathematical reasoning and exploration of potential solutions.
Participants appear to disagree on the nature of symmetry in the function and the implications for the maximal value. The discussion remains unresolved regarding the optimal approach to finding the maximal value.
There are limitations in the discussion regarding the assumptions made about symmetry and the specific values tested, which may affect the conclusions drawn about the maximal value.
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.
[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.kaliprasad said:Three triples are missing solution set is $(4,1,0),(4,0,1),(0,1,4), (0,4,1),(1,4,0),(1,0,4)$
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.
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