Let $a$, $b$, $c$ be positive reals satisfying $a + b + c = \sqrt[\large 7]{a} + \sqrt[\large 7]{b} + \sqrt[\large 7]{c}$. Prove that $a^a b^b c^c ≥ 1$.

This is the last problem from this excellent overview of various inequality problems by Evan Chen.

I suspect the author left it for the end of his document as a very difficult problem, or as a problem that has a beautiful or an unexpected solution.

I tried using Cauchy and Hoelder inequalities in some ways, and some substitutions, but no luck so far.

I am curious what would you guys say about the problem. Also, is there something special about number $7$ regarding this problem? Would the statement hold if there was $19$ instead of $7$? What about the case with four, five, six numbers? What about the case with only $a$ and $b$?

  • $\begingroup$ There was slight correction of the statement. The error was related to the editor on this site inserting some characters for quoted text.. Sorry about that. $\endgroup$
    – VividD
    Sep 5, 2015 at 16:42

2 Answers 2


First of all, the solution: by weighted AM-GM we have $$1 = \frac{a}{a+b+c} a^{-6/7} + \frac{b}{a+b+c} b^{-6/7} + \frac{c}{a+b+c} c^{-6/7} \ge (a^ab^bc^c)^{\frac{-6/7}{a+b+c}}. $$ Of course, from this solution it's clear that both the number of variables and the choice of the number $7$ is irrelevant. (Pedagogically and aesthetically, I'm of the opinion that it's usually better to include the specific, concrete instance if it's clear how to generalize it.)

Seeing as I'm the author of this problem (which was the second problem of ELMO 2013), I can comment on its creation too. I had long known that one could get $a^2+b^2+c^2 \ge a^ab^bc^c$, when $a+b+c=1$, using weighted AM-GM. I really wanted to see if I could get something more decent, since the problem is rather trivial in that formulation. After two hours of playing around with this weighted idea before I decided to put a condition of $a^2 + b^2 + c^2 = a + b + c$ so that the left-hand side of $\frac{a^2+b^2+c^2}{a+b+c} \ge a^{\frac{a}{a+b+c}} b^{\frac{b}{a+b+c}} c^{\frac{c}{a+b+c}}$ would be nice. Suddenly I realized I could just factor out the exponent, and the whole thing simplified super nicely as $a^ab^bc^c \le 1$. And so after some cosmetic changes it became the problem you see now.


Let $r> 1$ and $a+b+c = \sqrt[r]a+\sqrt[r]b+\sqrt[r]a$ for positive $a, b, c$. We need to show $a^ab^bc^c \ge 1$. Equivalently, it is enough to show for $x> 0$, $$f(x) = x \log x -\frac{r}{r-1}(x - \sqrt[r]x) \ge 0$$

Looking at $f'(x)$, we get that its sign is determined by the sign of $$g(x) = (r-1)\log x+\frac1{\sqrt[r]{x^{r-1}}}-1$$ and $g'(x) = \dfrac{(r-1)}{rx^2}(rx-\sqrt[r]x)$ which crosses the positive x-axis only once from the bottom. So $g(x)$ has two zeros, easily $g(1)=0$ and there is another in $(0, 1)$. The sign of $g(x)$ is then $+, - , +$ where the sign changes happen at the roots.

Finally, this means $f(x)$ has two extrema, the first is in $(0, 1)$ and is a maximum, and the second is at $x=1$, and is the minimum. As $f(1) = 0$ is this minimum, we have $f(x) \ge 0$

  • $\begingroup$ Thanks, I am just not clear about the equivalence of the initial problem and the first functional inequality. Would you mind explaining the details? $\endgroup$
    – VividD
    Sep 5, 2015 at 19:56
  • 1
    $\begingroup$ It is not equivalent, but one way implication. $f(x)\ge 0 \implies f(a)+f(b)+f(c)\ge 0 \implies a\log a + b\log b+ c\log c \ge 0 \implies $ the initial problem holds. This is because the portion in brackets sums to the constraint. Try it out and it should be evident. The method works for more variables as well. $\endgroup$
    – Macavity
    Sep 6, 2015 at 2:23

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