\(x^2+2x+1+y^2=\left(x+1\right)^2+y^2\)
Do \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\\y^2\ge0\end{matrix}\right.\) \(\forall x;y\)
\(\Rightarrow\left(x+1\right)^2+y^2\ge0\)
Dấu "=" vẫn xảy ra tại \(\left\{{}\begin{matrix}x=-1\\y=0\end{matrix}\right.\)