Problem
ALG-B1-M07-P009 Sum of pairwise products
#9
★★☆☆☆ Level 2 of 5
Let \(a,b,c\ge0\) and \(a+b+c=1\). Prove that \(ab+bc+ca\le\frac{1}{3}\).
Use \((a+b+c)^2\ge3(ab+bc+ca)\).
From the three-squares inequality, \((a+b+c)^2\ge3(ab+bc+ca)\). Since \(a+b+c=1\), we get \(1\ge3(ab+bc+ca)\), so \(ab+bc+ca\le\frac{1}{3}\). Equality holds at \(a=b=c=\frac{1}{3}\).
Shows how a basic sum of squares gives a sharp estimate.