Problem 625

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,
$$\min_{k=1,\dots,n} (x_k^2 + x_{k+1}^2) \le x_j^2 + x_{j+1}^2 \le 2x_{j+1}^2 \le 2x_{j+1}x_{j+2} \le \max_{k=1,\dots,n} (2x_k x_{k+1}).$$