Problem
ALG-B2-M07-P010 Nesbitt after normalization
#10
★★★☆☆ 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. Set \(a+b+c=1\), or apply Cauchy directly.
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 \((a+b+c)^2\ge3(ab+bc+ca)\), we get the lower bound \(\frac32\).
A classical degree-zero problem.