\( x^2+5y^2+2xy-4x-8y+2021\\=x^2+2xy+y^2-4x-4y+4y^2-4y+1+4+2016\\=(x+y)^2-4(x+y)+4+(2y-1)^2+2016\\=(x+y-2)^2+(2y-1)^2+2016\\Vì\,\,\begin{cases}(x+y-2)^2\ge 0\\(2y-1)^2\ge 0\end{cases}\\→(x+y-2)^2+(2y-1)^2+2016\ge 2016\\→\min =2016\\→Dấu\,\,bằng\,\,xảy\,\,ra\,\,khi\,\,\begin{cases}x+y-2=0\\2y-1=0\end{cases}\\↔\begin{cases}x=\dfrac{3}{2}\\y=\dfrac{1}{2}\end{cases}\\Vậy\,\,\min =2016\,\,khi\,\,x=\dfrac{3}{2};y=\dfrac{1}{2}\)