By the AM-GM inequality, we know that
$$\sqrt{(c-1) \cdot 1} \le \frac{c-1+1}{2},$$
therefore
$$c - 1 \le \frac{c^2}{4}$$
with equality for $c=2$. With the two analogous inequalities for $a$ and $b$ we obtain
$$\frac{ab}{c-1} + \frac{bc}{a-1} + \frac{ca}{b-1} \ge \frac{4ab}{c^2} + \frac{4bc}{a^2} + \frac{4ca}{b^2} \ge 12 \sqrt[3]{\frac{ab}{c^2} \cdot \frac{bc}{a^2} \cdot \frac{ca}{b^2}} = 12$$
where the last inequality is the AM-GM inequality again. Therefore, equality holds for $a = b = c = 2$.