Ta có :
\(2x^2+\frac{y^2}{4}+\frac{1}{x^2}=4\)
\(\Leftrightarrow\left(x^2-2+\frac{1}{x^2}\right)+\left(x^2+xy+\frac{y^2}{4}\right)=2+xy\)
\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(x+\frac{y}{2}\right)^2=2+xy\)
VT \(\ge\)0 \(\Rightarrow xy\ge-2\)
Dấu " = " xảy ra khi \(x=\frac{1}{x}=\frac{-y}{2}\Leftrightarrow\left\{{}\begin{matrix}x^2=1\\y=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(A_{min}\)= 2016-2=2014 khi \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
+, Ta có :
\(2x^2+\frac{y^2}{4}+\frac{1}{x^2}=4\)
\(\Leftrightarrow\left(x^2-2+\frac{1}{x^2}\right)+\left(x^2-xy+\frac{y^2}{4}\right)=2-xy\)
\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(x-\frac{y}{2}\right)^2=2-xy\)
VT \(\ge0\Rightarrow xy\le2\)
Dấu "=" xảy ra khi \(x=\frac{1}{x}=\frac{y}{2}\Leftrightarrow\left\{{}\begin{matrix}x^2=1\\y=2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(A_{max}\)= 2016+2=2018 khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\)
