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pab+qbc+rca <= /8(a + b + c)^2 if p, q, r \in [0,1/2], p + q + r = 1, a,b,c>0

Source: 2013 Moldova JBMO TST p5

February 25, 2021
inequalitiesalgebra

Problem Statement

The real numbers a,b,ca, b, c are positive, and the real numbers p,q,r∈[0,1/2]p, q, r \in [0,1/2] satisfy equality p+q+r=1p + q + r = 1. Prove the inequality pab+qbc+rca≤18(a+b+c)2.pab + qbc + rca \le \frac18 (a + b + c)^2.
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