jmc

Find the common denominator and replace the $ab$ in the numerator with $a-b$ to get

$$\begin{array}{rl}\frac{a}{b}+\frac{b}{a}-ab& =\frac{a^2+b^2-(ab{)}^2}{ab}\\ & =\frac{a^2+b^2-(a-b{)}^2}{ab}\\ & =\frac{a^2+b^2-(a^2-2ab+b^2)}{ab}\\ & =\frac{2ab}{ab}=2.\end{array}$$Hence the minimum possible value is the only possible value, $\boxed{2}$.