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Russian Mathematical Olympiad

Algebra Difficulty 5.1 AIME, harder Prove it Russia

Let $a$ , $b$ be real numbers such that $a^3 - b^3 = 2$ and $a^5 - b^5 \ge 4$ . Prove that $a^2 + b^2 \ge 2$ . (I. Bogdanov)

Числа $a$ и $b$ таковы, что $a^3 - b^3 = 2$ , $a^5 - b^5 \ge 4$ . Докажите, что $a^2 + b^2 \ge 2$ . (И. Богданов)

Solution

Заметим, что $2(a^2 + b^2) = (a^2 + b^2)(a^3 - b^3) = (a^5 - b^5) + a^2b^2(a - b) \ge 4 + a^2b^2(a - b)$ . Поскольку $a^3 > b^3$ , мы имеем $a > b$ , а значит, $a^2b^2(a - b) \ge 0$ . Итак, $2(a^2 + b^2) \ge 4$ , откуда и следует утверждение задачи.

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