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nice inequality by panaitopol

Source: JBMO 2002, Problem 4

September 21, 2003
inequalitiesfunctionrearrangement inequalityinequalities solved3-variable inequalitycyclic inequality

Problem Statement

Prove that for all positive real numbers a,b,ca,b,c the following inequality takes place 1b(a+b)+1c(b+c)+1a(c+a)≥272(a+b+c)2. \frac{1}{b(a+b)}+ \frac{1}{c(b+c)}+ \frac{1}{a(c+a)} \geq \frac{27}{2(a+b+c)^2} . Laurentiu Panaitopol, Romania
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