Problem
ALG-B2-M03-P020 Four parts in the denominator
#20
★★★★★ Level 5 of 5
Prove for \(a,b,c>0\): \[\frac{a^2}{2a+b+c}+\frac{b^2}{2b+c+a}+\frac{c^2}{2c+a+b}\ge\frac{a+b+c}{4}.\]
Hint 1. Add the denominators.
Hint 2. The denominator sum is \(4(a+b+c)\).
By Cauchy, \[\sum\frac{a^2}{2a+b+c}\ge\frac{(a+b+c)^2}{(2a+b+c)+(2b+c+a)+(2c+a+b)}=\frac{(a+b+c)^2}{4(a+b+c)}=\frac{a+b+c}{4}.\]
Cauchy-Schwarz module training problem. Method tags: cauchy, bounds.