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Romanian Mathematical Olympiad

Algebra Difficulty 5.1 AIME, harder Prove it Romania

Let $a, b, c$ be real numbers such that:
$|a - b| \ge |c|, \quad |b - c| \ge |a|, \quad |c - a| \ge |b|.$

Prove that one of the numbers $a, b, c$ equals the sum of the other two.

Solution

Squaring the first inequality gives $(a-b)^2 \ge c^2$ , hence $(a-b+c)(b+c-a) \ge 0$ . Multiplying the latter with the other two similar inequalities implies $(a+b-c)^2(b+c-a)^2(c+a-b)^2 \le 0$ , hence one of $a, b, c$ is the sum of the other two.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.