+ ĐK : \(x\ne-1\)
\(\frac{2x^2+2}{\left(x+1\right)^2}=\frac{\left(x^2+2x+1\right)+\left(x^2-2x+1\right)}{\left(x+1\right)^2}=\frac{\left(x+1\right)^2+\left(x-1\right)^2}{\left(x+1\right)^2}=1+\frac{\left(x-1\right)^2}{\left(x+1\right)^2}\ge1\forall x\ne-1\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)