Do \(x;y\) cùng dấu suy ra xy > 0
\(A=x^2+y^2+\dfrac{2}{xy}\ge2xy+\dfrac{2}{xy}\ge2\sqrt{2xy.\dfrac{2}{xy}}=4\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=y\\2xy=\dfrac{2}{xy}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=1\\x=y=-1\end{matrix}\right.\)