Problem
ALG-B2-M08-P018 Triangle sum of ratios
#18
★★★★★ Level 5 of 5
For triangle sides, prove \[\frac{a^2}{(b+c-a)^2}+\frac{b^2}{(c+a-b)^2}+\frac{c^2}{(a+b-c)^2}\ge3.\]
Hint. First prove the stronger-looking \(\sum \frac{a}{b+c-a}\ge3\), then use Cauchy or AM-GM.
Let \(u=\frac{a}{b+c-a}\), \(v=\frac{b}{c+a-b}\), \(w=\frac{c}{a+b-c}\). By Problem 6, \(u+v+w\ge3\). Then \(u^2+v^2+w^2\ge\frac{(u+v+w)^2}{3}\ge3\).
Teaching goal: recognize the appropriate substitution and check the domain.