Problem
ALG-B2-M11-P001 Nesbitt as Cauchy
#1
★★★☆☆ Level 3 of 5
Prove for \(a,b,c>0\): \[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac32.\]
Hint. Write \(\frac{a}{b+c}\) as \(\frac{a^2}{a(b+c)}\).
By Cauchy, \[\sum\frac{a}{b+c}=\sum\frac{a^2}{a(b+c)}\ge\frac{(a+b+c)^2}{2(ab+bc+ca)}.\] Since \((a+b+c)^2\ge3(ab+bc+ca)\), the lower bound is \(3/2\).
Teaching goal: the student should first recognise the type of estimate, then choose the tool.