Problem
ALG-B2-M10-P005 Squares in denominators
#5
★★★★★ Level 5 of 5
Prove for \(a,b,c>0\): \[\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2}\ge\frac32.\]
Hint. Cauchy gives a fraction with \((a+b+c)^2\), then use \( (a+b+c)^2\ge3(a^2+b^2+c^2)\).
By Cauchy, \[\sum\frac{a^2}{b^2+c^2}\ge\frac{(a+b+c)^2}{2(a^2+b^2+c^2)}.\] Since \((a+b+c)^2\ge3(a^2+b^2+c^2)\), the lower bound is \(3/2\).
Teaching goal: choose a combination of methods and always check the equality case.