Maximize the Cyclic sum

Let $x_1,x_2,\dots ,x_6$ be nonnegative real numbers such that $x_1+x_2+x_3+x_4+x_5+x_6=1$, and $x_1x_3x_5+x_2x_4x_6 \geq \frac{1}{540}$. Let $p$ and $q$ be positive relatively prime integers such that $\frac{p}{q}$ is the maximum possible value of $x_1x_2x_3+x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_6 + x_5x_6x_1 + x_6x_1x_2$. Find $p+q$. Hints only!

This was a very difficult problem actually. A possibility is: $x_k = \frac{1}{6}$ so that: $\sum_{cyc} x_1x_3x_5 = 1/108 > \frac{1}{504}$, which is a possiblity (true). I took: $540 = 5(3^3)(2^2)$ Obviously, $x_k < 1$ so, the bigger each number, the bigger the total max value. I would think $1/6$ is the best, to get: $6\cdot \frac{1}{216} = \frac{1}{36}$ But this seems way too easy! Hints only!

• ¿Where does this come from? Just curious... Aug 13, 2015 at 21:17
• @vonbrand, AIME II 2011 #9 Aug 13, 2015 at 21:36

To get an idea where to look for the maximum, note that without the inequality the maximum would be at $x_1=x_2=x_3=\frac13$, $x_4=x_5=x_6=0$, with a maximum value of $\frac1{27}$.