MathDB
Hard inequality

Source:

October 23, 2010
inequalities

Problem Statement

Given a,b,c a,b, c positive real numbers satisfying a+b+c=1 a+b+c=1 . Prove that 1ab+bc+ca2a3(b+c)+2b3(c+a)+2c3(a+b)a+b+c \dfrac{1}{\sqrt{ab+bc+ca}}\ge \sqrt{\dfrac{2a}{3(b+c)}} +\sqrt{\dfrac{2b}{3(c+a)}} + \sqrt{\dfrac{2c}{3(a+b)}} \ge \sqrt{a} +\sqrt{b}+\sqrt{c}