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An inequality where cyc \frac{1}{a} \ge \frac{3}{abc}

Source: Azerbaijan 2022 JBMO TST

July 3, 2022
inequalitiesAlgebra - InequalityalgebraJBMO TSTAzerbaijanJunior

Problem Statement

For positive real numbers a,b,ca,b,c, 1a+1b+1c≥3abc\frac{1}{a}+\frac{1}{b} + \frac{1}{c} \ge \frac{3}{abc} is true. Prove that: a2+b2a2+b2+1+b2+c2b2+c2+1+c2+a2c2+a2+1≥2 \frac{a^2+b^2}{a^2+b^2+1}+\frac{b^2+c^2}{b^2+c^2+1}+\frac{c^2+a^2}{c^2+a^2+1} \ge 2
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