jmc

We complete the square with respect to the terms $b^2$ and $\frac{b}{a},$ to get $$b^2+\frac{b}{a}={\left(b+\frac{1}{2a}\right)}^2-\frac{1}{4a^2}.$$This is minimized when $b=-\frac{1}{2a}.$ The problem now is to minimize $$a^2+\frac{1}{a^2}-\frac{1}{4a^2}=a^2+\frac{3}{4a^2}.$$We can assume that $a$ is positive. Then by AM-GM, $$a^2+\frac{3}{4a^2}\ge 2\sqrt{a^2\cdot \frac{3}{4a^2}}=\sqrt{3}.$$Equality occurs when $a=\sqrt[4]{\frac{3}{4}},$ so the minimum value is $\boxed{\sqrt{3}}.$