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Algebra Difficulty 4.7 AIME Prove it Ukraine

Let a,b[1,1]a, b \in [-1,1]. Prove that a1b2+b1a21a\sqrt{1-b^2} + b\sqrt{1-a^2} \le 1.

Solution

By substitution: a=sinαa = \sin \alpha, b=sinβb = \sin \beta the given inequality becomes: sinαcosβ+sinβcosα1\sin \alpha \cos \beta + \sin \beta \cos \alpha \le 1. This inequality holds for an arbitrary α,β\alpha, \beta.

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