2003 IMO 第 6 题

A6 (USA) Let $n$ be a positive integer and let $\left(x_{1}, \ldots, x_{n}\right),\left(y_{1}, \ldots, y_{n}\right)$ be two sequences of positive real numbers. Suppose $\left(z_{2}, z_{3}, \ldots, z_{2 n}\right)$ is a sequence of positive real numbers such that $$z_{i+j}^{2} \geq x_{i} y_{j} \quad \text { for all } 1 \leq i, j \leq n$$ Let $M=\max \left\{z_{2}, \ldots, z_{2 n}\right\}$. Prove that $$\left(\frac{M+z_{2}+\cdots+z_{2 n}}{2 n}\right)^{2} \geq\left(\frac{x_{1}+\cdots+x_{n}}{n}\right)\left(\frac{y_{1}+\cdots+y_{n}}{n}\right) .$$

A6 (美国) 令$n$为正整数，并令$\left(x_{1}, \ldots, x_{n}\right),\left(y_{1}, \ldots, y_{n}\right)$为两个正实数序列。假设 $\left(z_{2}, z_{3}, \ldots, z_{2 n}\right)$ 是正实数序列，使得 $$z_{i+j}^{2} \geq x_{i} y_{j} \quad \text { for all } 1 \leq i, j \leq n$$设 M=\max \left\{z_{2}, \ldots, z_{2 n}\右\}。证明 $$\left(\frac{M+z_{2}+\cdots+z_{2 n}}{2 n}\right)^{2} \geq\left(\frac{x_{1}+\cdots+x_{n}}{n}\right)\left(\frac{y_{1}+\cdots+y_{n}}{n}\right) 。$$