, 2016

Algebra Difficulty 5.0 AIME, harder Prove it Romania

Let $n$ be an odd positive integer, and let $x_1, x_2, \dots, x_n$ be non-negative real numbers. Show that $\min_{k=1,\dots,n} (x_k^2 + x_{k+1}^2) \le \max_{k=1,\dots,n} (2x_k x_{k+1})$ , where $x_{n+1} = x_1$ .

Solution

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

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.