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1 + 3/(a+b+c)>= 6/(ab+bc+ca) if a,b,c>0 with abc = 1

Source: 2003 Romania JBMO TST 2.1

June 1, 2020

algebrainequalities

Problem Statement

Let

a, b, c

be positive real numbers with

abc = 1

. Prove that

1 + \frac{3}{a+b+c}\ge \frac{6}{ab+bc+ca}