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(a + b + c)(ab + bc+ca) + 3abc>= 4(ab + bc + ca) if ab+bc+ca>= a+b+c

Source: 2014 Belarus TST 2.2

December 30, 2020

algebrainequalities

Problem Statement

Given positive real numbers

a,b,c

with

ab+bc+ca\ge a+b+c

, prove that

(a + b + c)(ab + bc+ca) + 3abc \ge 4(ab + bc + ca).

(I. Gorodnin)