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(1+a^2)(1+b^2)(1+c^2)/(1+a)(1+b)(1+c)>= \frac12 (1+abc) for a,b,c>=0

Source: Singapore Senior Math Olympiad 2008 2nd Round p5 SMO

March 25, 2020
inequalitiesalgebra

Problem Statement

Let a,b,c≥0a,b,c \ge 0. Prove that (1+a2)(1+b2)(1+c2)(1+a)(1+b)(1+c)≥12(1+abc)\frac{(1+a^2)(1+b^2)(1+c^2)}{(1+a)(1+b)(1+c)}\ge \frac12 (1+abc)
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