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Ukrainian Mathematical Olympiad

Algebra Difficulty 4.9 AIME Prove it Ukraine

Positive numbers $a$ and $b$ satisfy the equality $a + b + a^{-1} + b^{-1} = 5$ . Prove that $3\sqrt{a + b} \ge a + b + 2$ .

Solution

Let $u = a + b$ and $v = \frac{1}{a} + \frac{1}{b}$ . Then $u + v = 5$ . As is known, $uv \ge 4$ . Therefore, $5 = u + v \ge u + \frac{4}{u}$ . Hence, $u^2 - 5u + 4 \le 0$ . Thus, $1 \le \sqrt{u} \le 2$ , that is, $(\sqrt{u} - 1)(\sqrt{u} - 2) \le 0$ , and $3\sqrt{u} \ge u + 2$ .

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