Problem
ALG-B1-M07-P011 Nesbitt's inequality
#11
★★★☆☆ Level 3 of 5
Let \(a,b,c>0\). Prove \[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}.\]
Write \(\frac{a}{b+c}\) as \(\frac{a^2}{a(b+c)}\) and apply Cauchy.
By Cauchy, the sum is at least \(\frac{(a+b+c)^2}{a(b+c)+b(c+a)+c(a+b)}=\frac{(a+b+c)^2}{2(ab+bc+ca)}\).
Since \((a+b+c)^2\ge3(ab+bc+ca)\), we get \(\frac{(a+b+c)^2}{2(ab+bc+ca)}\ge\frac{3}{2}\). Equality holds when \(a=b=c\).
A standard problem, but it already requires two ideas: Cauchy and a symmetric estimate.