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Mongolian National Mathematical Olympiad

Algebra Difficulty 4.8 AIME Prove it Mongolia

Let $a$ and $b$ be real numbers such that $a + b = 1$ . Prove the following inequality.
$\sqrt{1 + 5a^2} + 5\sqrt{2 + b^2} \ge 9$

Solution

By the Cauchy-Schwarz inequality, we have
$3\sqrt{1 + 5a^2} = \sqrt{2^2 + (\sqrt{5})^2} \cdot \sqrt{1 + 5a^2} \geq 2 + 5a$
and
$3\sqrt{2 + b^2} = \sqrt{(2\sqrt{2})^2 + 1} \cdot \sqrt{2 + b^2} \geq 4 + b.$
Hence
$\sqrt{1 + 5a^2} + 5\sqrt{2 + b^2} \geq \frac{2 + 5a}{3} + \frac{5(4 + b)}{3} = 9.$
Equality holds for $a = b = 1/2$ .

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