Ta có $x+y+z=3 \Rightarrow xyz\geq1$(dễ dàng chứng minh)
$\frac{1}{x^{2}+x}+\frac{1}{y^{2}+y}+\frac{1}{z^{2}+z}\geq \frac{3}{\sqrt[3]{xyz(x+1)(y+1)(z+1)}}$(BĐT Côsi)
$\geq \frac{3}{\sqrt[3]{(x+1)(y+1)(z+1)}}\geq \frac{3}{\frac{x+1+y+1+z+1}{3}}=\frac{9}{6}=\frac{3}{2}$