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Moldova tst iii, problem 3

Source: Moldova TST III

March 26, 2006
inequalitiesinequalities proposed

Problem Statement

Positive real numbers a,b,ca,b,c satisfy the relation abc=1abc=1. Prove the inequality: a+3(a+1)2+b+3(b+1)2+c+3(c+1)23\frac{a+3}{(a+1)^{2}}+\frac{b+3}{(b+1)^{2}}+\frac{c+3}{(c+1)^{2}}\geq3.