Given $0<a<1,0<b<1$, and $a b=\frac{1}{36}$. Find the minimum value of $u=\frac{1}{1-a}+\frac{1}{1-b}$.
1.  By the AM-GM inequality $u \geqslant \frac{2}{\sqrt{(1-a)(1-b)}} \geqslant \frac{2}{\frac{1-a+1-b}{2}}=$ $\frac{4}{2-(a+b)} \geqslant \frac{4}{2-2 \sqrt{a b}}=\frac{4}{2-\frac{1}{3}}=\frac{12}{5}$. When $a=b=\frac{1}{6}$, the equality holds, hence the minimum value of $u$ is $\frac{12}{5}$.