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_{1} x_{3} x_{5}+x_{2} x_{4} x_{6} \geq \frac{1}{540}. Let p and q be positive relatively prime integers such
that \frac{p}{q} is the maximum possible value of
x_{1} x_{2} x_{3}+x_{2} x_{3} x_{4}+x_{3} x_{4} x_{5}+x_{4} x_{5} x_{6}+x_{5} x_{6} x_{1}+x_{6} x_{1} x_{2}. Find p+q.