Ta có:
$A=2+x+y+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{x}{y}+\dfrac{y}{x}$
$\geq 2+x+y+\dfrac{1}{x}+\dfrac{1}{y}+2\sqrt{\dfrac{x}{y}.\dfrac{y}{x}}$
$=4+2x+\dfrac{1}{x}+2y+\dfrac{1}{y}-(x+y)$
$\ge 4+2\sqrt{2x.\dfrac{1}{x}}+2\sqrt{2y.\dfrac{1}{y}}-\sqrt{2(x^2+y^2)}$
$=4+3\sqrt2$
$\min A=4+3\sqrt2 \Leftrightarrow x=y=\dfrac{1}{\sqrt2}$