Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Prove it Ukraine

What smallest value can be attained by the expression

$\frac{(x + y + |x - y|)^2}{xy}$

for positive $x$ , $y$ ?

Solution

We assume $x \ge y$ , then we can rewrite as:
$\frac{(x + y + |x - y|)^2}{xy} = \frac{(x + y + x - y)^2}{xy} = \frac{(2x)^2}{xy} = \frac{4x^2}{xy} = \frac{4x}{y} \ge 4$
When $x = y$ , we get equality.

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