Ta có $xy \le \frac{(x+y)^2}{4}=\frac14 \Rightarrow \frac3{8xy} \ge \frac32..$
Áp dụng BĐT Cô-si
$A = 2xy + \frac1{2xy} = 2xy + \frac1{8xy}+\frac3{8xy} \ge 2\sqrt{2xy . \frac1{8xy}} + \frac32 = \frac{5}2$.
Vậy $\min A =\frac{5}2 \Leftrightarrow x=y=\frac12$.