2012 USAMO 第 6 题

For integer $n \ge 2$, let $x_1$, $x_2$, $\dots$, $x_n$ be real numbers satisfying

$$x_1 + x_2 + \dots + x_n = 0, \quad \text{and} \quad x_1^2 + x_2^2 + \dots + x_n^2 = 1.$$

For each subset $A \subseteq \{1, 2, \dots, n\}$, define

$$S_A = \sum_{i \in A} x_i.$$

(If $A$ is the empty set, then $S_A = 0$.)

Prove that for any positive number $\lambda$, the number of sets $A$ satisfying $S_A \ge \lambda$ is at most $2^{n - 3}/\lambda^2$. For what choices of $x_1$, $x_2$, $\dots$, $x_n$, $\lambda$ does equality hold?

对于整数$n \ge 2$，设$x_1$、$x_2$、$\dots$、$x_n$为实数，满足

$$

x_1 + x_2 + \dots + x_n = 0，\quad \text{和} \quad x_1^2 + x_2^2 + \dots + x_n^2 = 1。

$$

对于每个子集 $A \subseteq \{1, 2, \dots, n\}$，定义

$$S_A = \sum_{i \in A} x_i。$$

(如果 $A$ 是空集，则 $S_A = 0$。)

证明对于任意正数$\lambda$，满足$S_A \ge \lambda$的集合$A$的个数至多为$2^{n - 3}/\lambda^2$。对于 $x_1$、$x_2$、$\dots$、$x_n$、$\lambda$ 的哪些选择，等式成立？