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Harvard-MIT Mathematics Tournament

Algebra Difficulty 5.0 AIME Prove it United States

Problem:
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 $x y z$ ?

Solution

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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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.