Prove that note, $\log _{b} a, \log _{a} b, \log _{a} c$ are all greater than 0, and their product equals 1. Thus, by the AM-GM inequality, we have
$$\begin{aligned} & \frac{\log _{b} a}{a+b}+\frac{\log _{c} b}{b+c}+\frac{\log _{a} c}{c+a} \geqslant 3 \sqrt[3]{\frac{\log _{b} a \cdot \log _{c} b \cdot \log _{a} c}{(a+b)(b+c)(c+a)}} \\ = & \frac{3}{\sqrt[3]{(a+b)(b+c)(c+a)}} \geqslant \frac{9}{(a+b)+(b+c)+(c+a)}=\frac{1}{2} \cdot \frac{9}{a+b+c}, \end{aligned}$$

Therefore, $2\left(\frac{\log _{b} a}{a+b}+\frac{\log _{c} b}{b+c}+\frac{\log _{a} c}{c+a}\right) \geqslant \frac{9}{a+b+c}$.