đặt $\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z \Rightarrow xyz=1$ VT=$\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y}\geq \frac{(x+y+z)^{2}}{2(x+y+z)}=\frac{x+y+z}{2}$
$\geq \frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}$
dấu "=" $\Leftrightarrow a=b=c=1$