\(x^2+2y^2+2xy+2y+2020\)
\(=\left(x^2+2xy+y^2\right)+\left(y^2+2y+1\right)+2019\)
\(=\left[\left(x+y\right)^2+\left(y+1\right)^2+2019\right]\ge2019\)
Vì \(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\forall x,y\\\left(y+1\right)^2\ge0\forall y\end{matrix}\right.\)
Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)