Question

对 $1 \le i \le 215$ ，令 $a_i = \frac{1}{2^i}$ ，并令 $a_{216} = \frac{1}{2^{215}}$ 。设 $x_1, x_2, \ldots, x_{216}$ 是正实数，满足 $\sum_{i=1}^{216} x_i = 1$ 以及 $\begin{aligned} &\sum_{1 \le i \lt j \le 216} x_i x_j \\ &= \frac{107}{215} \\ &\quad {}+ \sum_{i=1}^{216} \frac{a_i x_i^2}{2(1 - a_i)}. \end{aligned}$ $x_2 = \frac{m}{n}$ 是其最大可能值，其中 $m$ 与 $n$ 是互质的正整数。求 $m + n$ 。

For $1 \le i \le 215$ let $a_i = \frac{1}{2^i}$ and $a_{216} = \frac{1}{2^{215}}.$ Let $x_1, x_2, \ldots, x_{216}$ be positive real numbers such that $\sum_{i=1}^{216} x_i = 1$ and $\begin{aligned} &\sum_{1 \le i \lt j \le 216} x_i x_j \\ &= \frac{107}{215} \\ &\quad {}+ \sum_{i=1}^{216} \frac{a_i x_i^2}{2(1 - a_i)}. \end{aligned}$ The maximum possible value of $x_2 = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：863

知识点：柯西-施瓦茨不等式不等式代数变形

难度评级：3500

解答：

因为 $\sum x_i = 1$ ，所以 $2\sum_{i \lt j} x_i x_j = 1 - \sum x_i^2$ 。将题设等式乘以两倍并整理， $\begin{aligned} 1 - \sum_{i=1}^{216} x_i^2 &= \frac{214}{215} \\ &\quad {}+ \sum_{i=1}^{216} \frac{a_i x_i^2}{1 - a_i}, \end{aligned}$ 所以 $\begin{aligned} \frac{1}{215} &= \sum_{i=1}^{216}\left(1 + \frac{a_i}{1 - a_i}\right)x_i^2 \\ &= \sum_{i=1}^{216} \frac{x_i^2}{1 - a_i}. \end{aligned}$

现在 $\sum a_i = \left(\frac{1}{2} + \cdots + \frac{1}{2^{215}}\right)$ $+ \frac{1}{2^{215}} = 1$ ，所以 $\sum (1 - a_i) = 216 - 1 = 215$ 。由柯西-施瓦茨不等式， $\begin{aligned} 1 &= \left(\sum_{i=1}^{216} x_i\right)^2 \\ &\le \left(\sum_{i=1}^{216} \frac{x_i^2}{1 - a_i}\right) \\ &\quad {}\cdot \left(\sum_{i=1}^{216} (1 - a_i)\right) \\ &= \frac{1}{215} \cdot 215 = 1. \end{aligned}$

等号成立，所以 $x_i$ 与 $1 - a_i$ 成比例，迫使 $x_i = \frac{1 - a_i}{215}$ 。因此 $x_2$ 唯一可能、也就是最大可能的值为 $\frac{1 - \frac{1}{4}}{215} = \frac{3}{860}$ ，所以 $m + n = 3 + 860 = 863$ 。

Since $\sum x_i = 1,$ we have $2\sum_{i \lt j} x_i x_j = 1 - \sum x_i^2.$ Doubling the given equation and rearranging, $\begin{aligned} 1 - \sum_{i=1}^{216} x_i^2 &= \frac{214}{215} \\ &\quad {}+ \sum_{i=1}^{216} \frac{a_i x_i^2}{1 - a_i}, \end{aligned}$ so $\begin{aligned} \frac{1}{215} &= \sum_{i=1}^{216}\left(1 + \frac{a_i}{1 - a_i}\right)x_i^2 \\ &= \sum_{i=1}^{216} \frac{x_i^2}{1 - a_i}. \end{aligned}$

Now $\sum a_i = \left(\frac{1}{2} + \cdots + \frac{1}{2^{215}}\right)$ $+ \frac{1}{2^{215}} = 1,$ so $\sum (1 - a_i) = 216 - 1 = 215.$ By the Cauchy-Schwarz inequality, $\begin{aligned} 1 &= \left(\sum_{i=1}^{216} x_i\right)^2 \\ &\le \left(\sum_{i=1}^{216} \frac{x_i^2}{1 - a_i}\right) \\ &\quad {}\cdot \left(\sum_{i=1}^{216} (1 - a_i)\right) \\ &= \frac{1}{215} \cdot 215 = 1. \end{aligned}$

Equality holds, so $x_i$ is proportional to $1 - a_i,$ forcing $x_i = \frac{1 - a_i}{215}.$ The only, hence maximum, possible value of $x_2$ is $\frac{1 - \frac{1}{4}}{215} = \frac{3}{860},$ and $m + n = 3 + 860 = 863.$

2016 AIME II 第 15 题

答案：863

难度评级：3500

解答：

其他年份的第 15 题