AD svac-sơ có:
\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{b+c+a+c+a+b}=\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\)
Dấu "=" xảy ra <=> \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)
<=> a=b=c
Akai Haruma,Băng Băng 2k6