jmc

We can write $$a+\frac{1}{b(a-b)}=(a-b)+b+\frac{1}{b(a-b)}.$$By AM-GM, $$(a-b)+b+\frac{1}{b(a-b)}\ge 3\sqrt[3]{(a-b)b\cdot \frac{1}{b(a-b)}}=3.$$Equality occurs when $a=2$ and $b=1,$ so the minimum value is $\boxed{3}.$