Problem
ALG-B1-M07-P019 Hidden Nesbitt
#19
★★★★☆ Level 4 of 5
Let \(a,b,c>0\) and \(a+b+c=1\). Prove \[\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\ge\frac{3}{2}.\]
Replace \(1-a\) by \(b+c\), and similarly.
Since \(a+b+c=1\), we have \(1-a=b+c\), \(1-b=c+a\), \(1-c=a+b\). Thus the inequality becomes \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\), which is Nesbitt's inequality.
A hidden form of a familiar inequality.