Problem
ALG-B2-M07-P015 Fractions with opposite sums
#15
★★★★★ Level 5 of 5
Prove for \(a,b,c>0\): \[\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}.\]
Hint. Normalize \(a+b+c=1\), or apply Cauchy.
By Cauchy, \[\sum\frac{a^2}{b+c}\ge\frac{(a+b+c)^2}{(b+c)+(c+a)+(a+b)}=\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}.\]
One of the main homogeneous fractional forms.