HMMT February

Algebra Difficulty 5.0 AIME Find the answer

Let $x, y$ , and $z$ be positive real numbers such that $(x \cdot y)+z=(x+z) \cdot(y+z)$ . What is the maximum possible value of $x y z$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

The condition is equivalent to $z^{2}+(x+y-1) z=0$ . Since $z$ is positive, $z=1-x-y$ , so $x+y+z=1$ . By the AM-GM inequality, $x y z \leq\left(\frac{x+y+z}{3}\right)^{3}=\frac{1}{27}$ with equality when $x=y=z=\frac{1}{3}$ .

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