2020 USAMO 第 6 题

Let $n \ge 2$ be an integer. Let $x_1 \ge x_2 \ge \cdots \ge x_n$ and $y_1 \ge y_2 \ge \cdots \ge y_n$ be $2n$ real numbers such that

$$\begin{aligned} 0 &= x_1 + x_2 + \cdots + x_n = y_1 + y_2 + \cdots + y_n \\ \text{and }1 &= x_1^2+x_2^2+\cdots+x_n^2=y_1^2+y_2^2+\cdots+y_n^2. \end{aligned}$$

Prove that

$$

\sum_{i=1}^n(x_iy_i-x_iy_{n+1-i})\ge\frac{2}{\sqrt{n-1}}.

$$

令 $n \ge 2$ 为整数。设 $x_1 \ge x_2 \ge \cdots \ge x_n$ 和 $y_1 \ge y_2 \ge \cdots \ge y_n$ 为 $2n$ 实数，使得

$$

\begin{对齐} 0 &= x_1 + x_2 + \cdots + x_n = y_1 + y_2 + \cdots + y_n \\ \text{和 }1 &= x_1^2+x_2^2+\cdots+x_n^2=y_1^2+y_2^2+\cdots+y_n^2。 \结束{对齐}

$$

证明

$$\sum_{i=1}^n(x_iy_i-x_iy_{n+1-i})\ge\frac{2}{\sqrt{n-1}}。$$