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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The formula for finding the maximal value of a^4b+b^4c+c^4a is (a^4b+b^4c+c^4a)/3.
The variables in the formula for finding the maximal value are a, b, and c.
The maximal value is the largest possible value that can be obtained from the given expression. To find it, you can substitute different values for a, b, and c and compare the results.
Yes, the maximal value can be negative if the given variables and exponents result in a negative value when substituted into the formula.
Finding the maximal value is important in many scientific contexts, as it can help determine the most efficient or optimal solution in a given situation. It can also be used to analyze and compare data in various fields such as physics, chemistry, and economics.