# If $a^2 + b^2 + c^2 = 2$, find the maximum of $\prod(a^5+b^5)$

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.

• 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

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.

• "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).

Let $c=\min\{a,b,c\}$, $a^2+\frac{c^2}{2}=x$ and $b^2+\frac{c^2}{2}=y$.

Hence, $x+y=2$, $xy\leq1$, $\sqrt{x}+\sqrt{y}\leq\sqrt{(1+1)(x+y)}=2$ and

$a^5+c^5\leq\left(a^2+\frac{c^2}{2}\right)^{\frac{5}{2}}=x^{\frac{5}{2}}$,$b^5+c^5\leq\left(b^2+\frac{c^2}{2}\right)^{\frac{5}{2}}=y^{\frac{5}{2}}$ and $$a^5+b^5\leq\left(a^2+\frac{c^2}{2}\right)^{\frac{5}{2}}+\left(b^2+\frac{c^2}{2}\right)^{\frac{5}{2}}=x^{\frac{5}{2}}+y^{\frac{5}{2}}.$$ Thus, $$\prod_{cyc}(a^5+b^5)\leq(xy)^{\frac{5}{2}}\left(x^{\frac{5}{2}}+y^{\frac{5}{2}}\right)=(\sqrt{x}+\sqrt{y})(xy)^{\frac{5}{2}}(x^2-\sqrt{x^3y}+xy-\sqrt{xy^3}+y^2)\leq$$ $$\leq2(xy)\cdot(xy)\cdot(xy)^{\frac{1}{2}}(x^2+xy+y^2-2\sqrt{xy})\leq$$ $$\leq2(xy)\cdot(xy)^{\frac{1}{2}}\cdot(xy)^{\frac{1}{2}}(x^2+xy+y^2-2\sqrt{xy})\leq$$ $$\leq2\left(\frac{xy+\sqrt{xy}+\sqrt{xy}+x^2+xy+y^2-2\sqrt{xy}}{4}\right)^4=2\left(\frac{(x+y)^2}{4}\right)^4=2.$$