Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

2 Answers 2

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 artsofproblemsolving.com, inequality section.

For the statement and a few applications of this principle, see http://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf

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.

share|improve this answer
2  
"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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.