Problem
ALG-B1-M10-P015 Estimate of products
#15
★★★☆☆ Level 3 of 5
Let \(a,b,c\ge0\), \(a+b+c=3\). Prove that \(ab+bc+ca\le3\).
Use \((a+b+c)^2\ge3(ab+bc+ca)\).
\((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\ge3(ab+bc+ca)\). Since \(a+b+c=3\), \(9\ge3(ab+bc+ca)\), so \(ab+bc+ca\le3\).
A symmetric inequality in a mixed set.