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Given that $a, b, c > 0$ and $a^2 + b^2 + c^2 = 2$, what is the maximum value of $(a^5 + b^5)(a^5 + c^5)(b^5 + c^5)$?

Normally when I encounter a problem like this, I seem to be able to push through with AM-GM. This one seems a little problematic since I can't seem to make any valid substitutions. I would prefer an elementary solution if there is one, that is without resorting to calculus. Thanks for your help.

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I would expect this kind of problem to have some kind of symmetry or extremity to the solution, so one of (i) $a=b=c=\sqrt{2/3}$, (ii) $a=b=1$ and $c=0$, or (iii) $a=\sqrt{2}$ and $b=c=0$, or some permutation of these. It seems (ii) gives a maximum value (of 2) to the product, but I don't see an easy way of showing it. – Henry Jul 16 '11 at 20:56
if all else fails, try lagrange multipliers – yoyo Jul 16 '11 at 21:20
up vote 3 down vote accepted

A useful principle for solving this kind of problem is the (n-1)-Equal variable principle by Vasile Cirtoaje. Lots of examples can be found on, inequality section.

For the statement and a few applications of this principle, see

For this problem, apply corollary 1.4 of the above paper, subjecting to the condition: $a^2+b^2+c^2 = 2$, and $a^5+b^5+c^5 = C$, a constant. Next, check that derivative of $f(x) = ln(C - x^5)$ is strictly concave, and $-f$ satisfies the condition for corollary 1.4. Thus maximum of the product is achieved (i.e. minimum of sum of -f at a,b,c) when 2 of the $0 \leq c \leq b = a$. (WLOG, assume $a \ge b \ge c$ beforehand) The rest is an exercise in single variable calculus.

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"check that the derivative is strictly concave" = "resorting to calculus" :-) – joriki Jul 16 '11 at 21:42
Agree. However, this theorem usually gives far less computations comparing to Lagrange multiplier in my experience. Also, only single-variable calculus is involved. – Soarer Jul 16 '11 at 21:44

(Not really an answer, just an observation). The problem is equivalent to proving the following, for any nonnegative $a,b,c$:

$$ \left[ \frac{(a^5+b^5)(b^5+c^5)(c^5+a^5)}{2} \right]^{1/15} \le \left[ \frac{a^2 + b^2+ c^2}{2} \right]^{1/2} $$

with equality attained for $a=b,c=0$ (or the symmetric equivalent alternatives). This kind of symmetric inequalities is typical in math olympiads, and there is much material, resources and tricks (but no easy recipes).

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