Giải:
Ta có: \(x^2+y^2\le2x+4y\)
\(\Leftrightarrow x^2+y^2-2x-4y\le0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)\le5\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2\le5\)
Mặt \(e\): \(A=2x+y=2x-2+y-2+4=2\left(x-1\right)+y-2+4\)
\(\Rightarrow A-4=2\left(x-1\right)+y-2\)
Áp dụng Bunhia có:
\(\left(A-4\right)^2=\left[2\left(x-1\right)+y-2\right]^2\le\left(2^2+1^2\right)\left[\left(x-1\right)^2+\left(y-2\right)^2\right]\le25\)
\(\Rightarrow-5\le A-4\)
\(\Rightarrow A\ge-1\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
