Question

设 $x_1$ ， $x_2$ ， $\ldots$ ， $x_6$ 为非负实数，满足 $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 =$ $1$ ，且 $x_1x_3x_5 + x_2x_4x_6 \ge \frac{1}{540}$ 。设 $p$ 和 $q$ 为互质正整数，使得 $\frac{p}{q}$ 是 $\begin{aligned} &x_1x_2x_3 + x_2x_3x_4 \\ &\quad {}+ x_3x_4x_5 + x_4x_5x_6 \\ &\quad {}+ x_5x_6x_1 + x_6x_1x_2. \end{aligned}$ 的最大可能值。求 $p + q$ 。

Let $x_1,$ $x_2,$ $\ldots,$ $x_6$ be nonnegative real numbers such that $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 =$ $1,$ and $x_1x_3x_5 + x_2x_4x_6 \ge \frac{1}{540}.$ Let $p$ and $q$ be positive relatively prime integers such that $\frac{p}{q}$ is the maximum possible value of $\begin{aligned} &x_1x_2x_3 + x_2x_3x_4 \\ &\quad {}+ x_3x_4x_5 + x_4x_5x_6 \\ &\quad {}+ x_5x_6x_1 + x_6x_1x_2. \end{aligned}$ Find $p + q.$

答案

答案：559

知识点：算术-几何平均不等式因式分解极限情形界定

难度评级：3060

解答：

令 $r = x_1x_3x_5 + x_2x_4x_6$ ，并令 $s$ 为题目中的循环和。展开 $(x_1 + x_4)(x_2 + x_5)(x_3 + x_6)$ 会得到八个三项乘积，正好是 $s$ 的六项加上 $r$ 的两项。因此 $r + s =$ $(x_1 + x_4)(x_2 + x_5)(x_3 + x_6)$ ，由 AM-GM 可知它至多为 $\left(\frac{1}{3}\right)^3 = \frac{1}{27}$ 。

因此 $\begin{aligned} s &\le \frac{1}{27} - r \\ &\le \frac{1}{27} - \frac{1}{540} \\ &= \frac{20 - 1}{540} = \frac{19}{540}. \end{aligned}$ 等号需要 $x_1 + x_4 = x_2 + x_5 = x_3 + x_6 =$ $\frac{1}{3}$ ，且 $r = \frac{1}{540}$ ：取 $x_1 = x_3 = \frac{3}{10}$ ， $x_5 = \frac{1}{60}$ ， $x_2 = \frac{19}{60}$ ， $x_4 = x_6 = \frac{1}{30}$ 。于是 $r = \frac{9}{6000} + \frac{19}{54000} = \frac{100}{54000} = \frac{1}{540}$ ，符合要求。

所以最大值为 $\frac{19}{540}$ ， $p + q = 19 + 540 = 559$ 。

Let $r = x_1x_3x_5 + x_2x_4x_6$ and let $s$ be the cyclic sum in question. Expanding $(x_1 + x_4)(x_2 + x_5)(x_3 + x_6)$ produces eight triple products, which are exactly the six terms of $s$ together with the two terms of $r.$ So $r + s =$ $(x_1 + x_4)(x_2 + x_5)(x_3 + x_6),$ and by AM-GM this is at most $\left(\frac{1}{3}\right)^3 = \frac{1}{27}.$

Therefore $\begin{aligned} s &\le \frac{1}{27} - r \\ &\le \frac{1}{27} - \frac{1}{540} \\ &= \frac{20 - 1}{540} = \frac{19}{540}. \end{aligned}$ Equality needs $x_1 + x_4 = x_2 + x_5 = x_3 + x_6 =$ $\frac{1}{3}$ with $r = \frac{1}{540}:$ take $x_1 = x_3 = \frac{3}{10},$ $x_5 = \frac{1}{60},$ $x_2 = \frac{19}{60},$ $x_4 = x_6 = \frac{1}{30}.$ Then $r = \frac{9}{6000} + \frac{19}{54000} = \frac{100}{54000} = \frac{1}{540},$ as required.

So the maximum is $\frac{19}{540},$ and $p + q = 19 + 540 = 559.$

2011 AIME II 第 9 题

答案：559

难度评级：3060

解答：

其他年份的第 9 题