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Inequality with a+b+c = 1 [Serbia MO 2017, D1, P1]

Source: Serbia National Mathematical Olympiad 2017, Day 1, Problem 1

April 1, 2017
inequalities

Problem Statement

Prove that for positive real numbers a,b,ca,b,c such that a+b+c=1a+b+c=1, a2b+1+b2c+1+c2a+12(a2+b2+c2).a\sqrt{2b+1}+b\sqrt{2c+1}+c\sqrt{2a+1}\le \sqrt{2-(a^2+b^2+c^2)}.