Example 11 Let real numbers $a, b$ satisfy $0<a<1,0<b<1$, and $ab=\frac{1}{36}$, find the minimum value of $u=\frac{1}{1-a}+\frac{1}{1-b}$. (Example 4 from [2])
From the condition, we know that $a+b \geqslant 2 \sqrt{a b}=\frac{1}{3}, u=\frac{1}{1-a}+$ $\frac{1}{1-b}=\frac{1^{2}}{1-a}+\frac{1^{2}}{1-b} \geqslant \frac{(1+1)^{2}}{1-a+1-b}=\frac{4}{2-(a+b)} \geqslant$ $\frac{12}{5}$. The equality holds if and only if $\frac{1}{1-a}=\frac{1}{1-b}$ and $a+b=\frac{1}{3}$, i.e., $a=b=$ $\frac{1}{6}$. Therefore, the minimum value sought is $\frac{12}{5}$.