Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

TRIVIAL. A formula, more than well-known and more than straightforward to prove, states that $1^2+2^2+...+n^2 = \frac13 n^3 + \frac12 n^2 + \frac16 n$. Hence, $\frac13 n^2 + \frac12 n + \frac16 = \frac{\frac13 n^3 + \frac12 n^2 + \frac16 n}{n} = \frac{1^2+2^2+...+n^2}{n}$. On the other hand, $\left(n!\right)^{\frac{2}{n}} = \sqrt[n]{\left(n!\right)^2} = \sqrt[n]{\left(1\cdot 2\cdot ...\cdot n\right)^2} = \sqrt[n]{1^2\cdot 2^2\cdot ...\cdot n^2}$. Hence, the inequality we have to prove transforms into $\frac{1^2+2^2+...+n^2}{n} \geq \sqrt[n]{1^2\cdot 2^2\cdot ...\cdot n^2}$, what is clear from the AM-GM inequality. Equality holds only if the numbers $1^2$, $2^2$, ..., $n^2$ are all equal, what is only possible for n = 1. Darij

Sorry for bumping an old thread, but I wanted to post another solution

We have that $1^2+2^2+...+n^2 = \frac13 n^3 + \frac12 n^2 + \frac16 n$, which can be proved here. Now $\frac13 n^2 + \frac12 n + \frac16 = \frac{1^2+2^2+...+n^2}{n}$, now by AM-GM (which can be proved here) we get that $\frac{1^2+2^2+...+n^2}{n} \ge \sqrt[n]{1^2\cdot 2^2\cdot ...\cdot n^2}.$ again sorry for bumping