Problem
ALG-B2-M03-P011 Squares over sums
#11
★★★★★ Level 5 of 5
Prove for \(a,b,c>0\): \[\frac{a^2}{b+2c}+\frac{b^2}{c+2a}+\frac{c^2}{a+2b}\ge\frac{a+b+c}{3}.\]
Hint 1. The denominator sum is \(3(a+b+c)\).
Hint 2. Engel form gives the desired estimate directly.
By Cauchy, \[\sum\frac{a^2}{b+2c}\ge\frac{(a+b+c)^2}{(b+2c)+(c+2a)+(a+2b)}=\frac{(a+b+c)^2}{3(a+b+c)}=\frac{a+b+c}{3}.\]
Cauchy-Schwarz module training problem. Method tags: cauchy, bounds.