\(2x^2+\dfrac{1}{x^2}+4y^2=4\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(x^2+\dfrac{1}{x^2}-2\right)+4xy=2\)
\(\Leftrightarrow\left(x-2y\right)^2+\left(x-\dfrac{1}{x}\right)^2+4xy=2\)
\(\Rightarrow4xy=2-\left(x-2y\right)^2-\left(x-\dfrac{1}{x}\right)^2\le2\)
\(\Rightarrow xy\le\dfrac{1}{2}\)
\(P_{max}=\dfrac{1}{2}\) khi \(\left\{{}\begin{matrix}x-2y=0\\x-\dfrac{1}{x}=0\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;\dfrac{1}{2}\right);\left(-1;-\dfrac{1}{2}\right)\)
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