Ta có: [tex]A=\frac{1}{x^2+y^2}+\frac{2}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{3}{2xy}+24xy-20xy\geq \frac{4}{(x+y)^2}+2\sqrt{\frac{3}{2xy}.24xy}-20(\frac{x+y}{2})^2=4+12-20.\frac{1}{4}=11[/tex]
Dấu "=" xảy ra khi [tex]x=y=\frac{1}{2}[/tex]
Vậy Min A = 11.