\(A=\dfrac{1}{a^2+b^2}+\dfrac{1}{ab}+4ab=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}+8ab-4ab\ge\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{1}{2}.8}-\dfrac{4.\left(a+b\right)^2}{4}=\dfrac{4}{\left(a+b\right)^2}+4-\left(a+b\right)^2\ge4+4-1=7\Rightarrow minA=7\Leftrightarrow a=b=\dfrac{1}{2}\)