# 2011 AIME II Problems/Problem 9

## Problem 9

Let $x_1, x_2, ... , x_6$ be non-negative real numbers such that $x_1 +x_2 +x_3 +x_4 +x_5 +x_6 =1$, and $x_1 x_3 x_5 +x_2 x_4 x_6 \ge {\frac{1}{540}}$. Let $p$ and $q$ be positive relatively prime integers such that $\frac{p}{q}$ is the maximum possible value of $x_1 x_2 x_3 + x_2 x_3 x_4 +x_3 x_4 x_5 +x_4 x_5 x_6 +x_5 x_6 x_1 +x_6 x_1 x_2$. Find $p+q$.

## Solution

Note that neither the constraint nor the expression we need to maximize involves products $x_i x_j$ with $i \equiv j \pmod 3$. Factoring out say $x_1$ and $x_4$ we see that the constraint is $x_1(x_3x_5) + x_4(x_2x_6) \ge {\frac1{540}}$, while the expression we want to maximize is $x_1(x_2x_3 + x_5x_6 + x_6x_2) + x_4(x_2x_3 + x_5x_6 + x_3x_5)$. Adding the left side of the constraint to the expression, we get: $(x_1 + x_4)(x_2x_3 + x_5x_6 + x_6x_2 + x_3x_5) = (x_1 + x_4)(x_2 + x_5)(x_3 + x_6)$. This new expression is the product of three non-negative terms whose sum is equal to 1. By AM-GM this product is at most $\frac1{27}$. Since we have added at least $\frac{1}{540}$ the desired maximum is at most $\frac1{27} - \frac1{540} =\frac{19}{540}$. It is easy to see that this upper bound can in fact be achieved by ensuring that the constraint expression is equal to $\frac1{540}$ with $x_1 + x_4 = x_2 + x_5 = x_3 + x_6 =\frac13$—for example, by choosing $x_1$ and $x_2$ small enough—so our answer is $540 + 19 = \fbox{559}.$

An example is: \begin{align*} x_3 &= x_6 = \frac16 \\ x_1 &= x_2 = \frac{5 - \sqrt{20}}{30} \\ x_5 &= x_4 = \frac{5 + \sqrt{20}}{30} \end{align*}

Another example is \begin{align*} x_1 = x_3 = \frac{1}{3} \\ x_2 = \frac{19}{60}, \ x_5 = \frac{1}{60} \\ x_4 &= x_6 = 0 \end{align*}

## Solution 2 (Not legit)

There's a symmetry between $x_1, x_3, x_5$ and $x_2,x_4,x_6$. Therefore, a good guess is that $a = x_1 = x_3 = x_5$ and $b = x_2 = x_4 = x_6$, at which point we know that $a+b = 1/3$, $a^3+b^3 \geq 1/540$, and we are trying to maximize $3a^2b+3ab^2$. Then, $$3a^3b+3ab^2 = (a+b)^3-a^3-b^3 \leq 1/27 - 1/540 = 19/540$$ which is the answer.

This solution is extremely lucky; if you attempt to solve for $a$ and $b$ you receive complex answers (which contradict the problem statement), but the final answer is correct.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 