The last (and the weirdest) problem from Chen`s "Brief Introduction to Olympiad Inequalities"

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 $f(x) \ge 0$