$A=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{4ab}+(4ab+\frac{1}{4ab})\geq \frac{4}{(a+b)^2}+\frac{1}{4(\frac{a+b}{2})^2}+2\sqrt{\frac{4ab}{2ab}}\geq 7$
$A=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{
5}{4ab}+(4ab+\frac{1}{4ab})\geq \frac{4}{(a+b)^2}+\frac{
5}{(a+b)^2}+2\sqrt{\frac{4ab}{
4ab}}\geq
4+5+2=11$Vậy $A{min}=11$ tại $a=b=\frac{1}{2}$