2. Let $a, b, c$ be positive numbers, then the minimum value of $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$ is $\qquad$ .
Let's assume $a \leq b \leq c$,
then $a+b \leq a+c \leq b+c, \frac{1}{a+b} \geq \frac{1}{a+c} \geq \frac{1}{b+c}$,
so $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$ is a sequence sum,
$$\begin{array}{l}
\therefore \frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a} \geq \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \\
\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a} \geq \frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}
\end{array}$$

Adding the two inequalities gives the minimum value of $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$ as $\frac{3}{2}$.