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!

  • $\begingroup$ ¿Where does this come from? Just curious... $\endgroup$
    – vonbrand
    Aug 13, 2015 at 21:17
  • $\begingroup$ @vonbrand, AIME II 2011 #9 $\endgroup$
    – Amad27
    Aug 13, 2015 at 21:36

1 Answer 1


There are two cases: Either the inequality is an equality, or you can ignore it. Accordingly, you have two constrained optimization problems, one with two constraints, one with one. For both you can write down the stationarity conditions with Lagrange parameters.

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 $\frac1{27}$.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ Aug 15, 2015 at 9:37

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