2011 AIME II Problem 9

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 relatively prime positive 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.

Leading zeroes must be inputted, so if your answer is 34, then input 034

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.