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\frac{a^3+a^2}{1+bc}+\frac{b^3+b^2}{1+ca}+\frac{c^3+c^2}{1+ab}\geq3.

Source: Moldova EGMO TST 2021

February 4, 2023
inequalities

Problem Statement

Postive real numbers a,b,ca, b, c satisfy abc=1abc=1. Show that a3+a21+bc+b3+b21+ca+c3+c21+ab3.\frac{a^3+a^2}{1+bc}+\frac{b^3+b^2}{1+ca}+\frac{c^3+c^2}{1+ab}\geq3.