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Ukrainian National Mathematical Olympiad - Third Round

Algebra Difficulty 4.7 AIME Prove it Ukraine

Prove the inequality for positive $a$ , $b$ , $c$ , $d$
$\frac{(a+b)^2}{cd} + \frac{(c+d)^2}{ab} \ge 8.$

Solution

Use two inequalities of means:
$\frac{(a+b)^2}{cd} + \frac{(c+d)^2}{ab} \ge \frac{4ab}{cd} + \frac{4cd}{ab} \ge 2 \cdot 4 \sqrt{\frac{ab}{cd} \cdot \frac{cd}{ab}} = 8.$

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