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Romania Junior TST 2021 Day 3 P1

Source:

June 7, 2021
algebrainequalitiesromaniaRomanian TST

Problem Statement

Let a,b,c>0a,b,c>0 be real numbers with the property that a+b+c=1a+b+c=1. Prove that 1a+bc+1b+ca+1c+ab71+abc.\frac{1}{a+bc}+\frac{1}{b+ca}+\frac{1}{c+ab}\geq\frac{7}{1+abc}.