Problem
ALG-B2-M02-P009 Three square roots
#9
★★★★☆ Level 4 of 5
Let \(a,b,c>0\), \(a+b+c=1\). Prove \[\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge\sqrt{\frac32}.\]
Hint 1. First apply Cauchy to the sum of square roots.
Hint 2. Then use \(\sum a(b+c)\le\frac{2}{3}(a+b+c)^2\).
By Cauchy, \[\left(\sum\sqrt{\frac{a}{b+c}}\right)^2\ge\frac{(a+b+c)^2}{a(b+c)+b(c+a)+c(a+b)}.\] Since the denominator is \(2(ab+bc+ca)\le\frac{2}{3}(a+b+c)^2=\frac23\), the sum is at least \(\sqrt{\frac32}\). This standard bound illustrates the same preparation principle.
AM-GM module training problem. Method tags: am-gm, bounds.