The discussion focuses on finding the maximal value of the expression \( a^4b + b^4c + c^4a \) under the constraint \( a + b + c = 5 \), where \( a, b, c \) are non-negative real numbers. It is established that the function does not exhibit complete symmetry among the variables, as demonstrated by the counterexamples \( f(4,0,1) \), \( f(1,4,0) \), and \( f(0,1,4) \), all yielding a value of 4. The conclusion emphasizes the importance of recognizing the cyclic symmetry in the variables while noting the lack of complete symmetry.
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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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