17. If $0<a, b, c<1$ satisfies $a b+b c+c a=1$, find the minimum value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$. (2004 Sichuan Province Mathematics Competition Problem)
17. Since $(a+b+c)^{2} \geqslant 3(a b+b c+c a)=3$, therefore, $a+b+c \geqslant \sqrt{3}$. By the Cauchy-Schwarz inequality, we have
$$\left(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\right)[(1-a)+(1-b)+(1-c)] \geqslant 9$$

Thus,
$$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geqslant \frac{9}{3-(a+b+c)} \geqslant \frac{9}{3-\sqrt{3}}=\frac{3(3+\sqrt{3})}{2}$$

When and only when $a=b=c=\frac{\sqrt{3}}{3}$, $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ achieves its minimum value $\frac{3(3+\sqrt{3})}{2}$.