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Romanian Mathematical Olympiad

Algebra Difficulty 5.0 AIME, harder Prove it Romania

The positive numbers aa, bb, cc are such that
ab+c+1+ba+c+1+ca+b+1≤1. \frac{a}{b+c+1} + \frac{b}{a+c+1} + \frac{c}{a+b+1} \le 1.
Prove that:
1b+c+1+1a+c+1+1a+b+1≥1. \frac{1}{b+c+1} + \frac{1}{a+c+1} + \frac{1}{a+b+1} \ge 1.

Solution

Denote by S=∑1b+c+1S = \sum \frac{1}{b+c+1}. Using the inequality from the hypothesis we obtain
∑(ab+c+1+1)≤4, \sum \left( \frac{a}{b+c+1} + 1 \right) \le 4,
and thus (a+b+c+1)S≤4(a+b+c+1)S \le 4.

Using arithmetic mean – harmonic mean inequality we get
S≥92(a+b+c+1)+1, S \ge \frac{9}{2(a + b + c + 1) + 1},
hence S≥9−2(a+b+c+1)S≥1S \ge 9 - 2(a + b + c + 1)S \ge 1.

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