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Algebra Difficulty 4.7 AIME Prove it Ukraine

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

Solution

Use two inequalities of means:
(a+b)2cd+(c+d)2ab≥4abcd+4cdab≥2⋅4abcd⋅cdab=8. \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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