jmc

Note that $$B^2={\left(x-\frac{1}{x}\right)}^2=x^2-2+\frac{1}{x^2}=A-2,$$so $$\frac{A}{B}=\frac{B^2+2}{B}=B+\frac{2}{B}.$$By AM-GM, $$B+\frac{2}{B}\ge 2\sqrt{B\cdot \frac{2}{B}}=2\sqrt{2}.$$Equality occurs when $x-\frac{1}{x}=\sqrt{2}$ (which has $x=\frac{1+\sqrt{3}}{\sqrt{2}}$ as a root), so the minimum value is $\boxed{2\sqrt{2}}.$