MathDB
abc >= (a+b-c)(b+c-a)(c+a-b) for a,b,c>0

Source: 1982 Swedish Mathematical Competition p2

March 28, 2021
algebrainequalities

Problem Statement

Show that abc(a+bc)(b+ca)(c+ab) abc \geq (a+b-c)(b+c-a)(c+a-b) for positive reals aa, bb, cc.