2016 European Girls' Mathematical Olympiad

In what follows, indices are reduced modulo $n$. Consider the $n$ differences $x_{k+1}-x_k$, $k=1,\dots ,n$. Since $n$ is odd, there exists an index $j$ such that $(x_{j+1}-x_j)(x_{j+2}-x_{j+1})\ge 0$. Without loss of generality, we may and will assume both factors non-negative, so $x_j\le x_{j+1}\le x_{j+2}$. Consequently, $$\underset{k=1,\dots ,n}{\min }(x_k^2+x_{k+1}^2)\le x_j^2+x_{j+1}^2\le 2x_{j+1}^2\le 2x_{j+1}{x}_{j+2}\le \underset{k=1,\dots ,n}{\max }(2x_k{x}_{k+1}).$$