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All-Soviet-Union Mathematical Olympiad

Algebra Difficulty 4.5 AIME Prove it Soviet Union

Problem:
Show that $\dfrac{(a + b)^2}{2} + \dfrac{a + b}{4} \geq a\sqrt{b} + b\sqrt{a}$ for all positive $a$ and $b$ .

Solution

Solution:
By AM/GM $\sqrt{ab} \leq \dfrac{a + b}{2}$ , so $\sqrt{2} (a + b) + \sqrt{ab} \leq a + b$ . Hence $\sqrt{2a + 2b} \geq \sqrt{a} + \sqrt{b}$ $\quad(*)$ .

By AM/GM $(a + b) \geq 2\sqrt{ab}$ and $2(a + b) + 1 \geq 2\sqrt{2a + 2b}$ . Multiplying, $(a + b)(2a + 2b + 1) \geq 4\sqrt{ab}\sqrt{2a + 2b}$ . Then using $(*)\geq 4\sqrt{ab}(\sqrt{a} +\sqrt{b})$ .

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