Austrian Mathematical Olympiad

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 $a=b=c=2$.