Ta có: \(4x^2+y^2=8+3xy\Leftrightarrow4x^2-4xy+y^2=8-xy\)
\(\Leftrightarrow\left(2x-y\right)^2=8-xy\ge0\forall x,y\inℝ\Rightarrow xy\le8\)
\(\Rightarrow P=xy+2020\le8+2020=2028\)
Đẳng thức xảy ra khi \(\hept{\begin{cases}2x=y\\xy=8\end{cases}}\Rightarrow\left(x,y\right)\in\left\{\left(2;4\right);\left(-2;-4\right)\right\}\)