Question

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 $x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_6 + x_5x_6x_1 + x_6x_1x_2.$ Find $p + q.$

Answer

Solution:

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 $s \le \frac{1}{27} - r \le \frac{1}{27} - \frac{1}{540} = \frac{20 - 1}{540} = \frac{19}{540}.$ 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 Problem 9

Solution:

Problem 9 in Other Years