Prove that the original inequality is equivalent to
$$\begin{aligned} & \frac{b}{a}+\frac{c}{b}+\frac{a}{c} \geqslant \frac{3}{2}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \\ \Leftrightarrow & \frac{b}{a}-\frac{b}{c+a}+\frac{c}{b}-\frac{c}{a+b}+\frac{a}{c}-\frac{a}{b+c} \geqslant \frac{3}{2} \\ \Leftrightarrow & \frac{b c}{a(c+a)}+\frac{c a}{b(a+b)}+\frac{a b}{c(b+c)} \geqslant \frac{3}{2} \end{aligned}$$

By the Cauchy-Schwarz inequality, we have
$$\begin{aligned} & {[(b+c)+(c+a)+(a+b)]\left[\frac{b c}{a(c+a)}+\frac{c a}{b(a+b)}+\frac{a b}{c(b+c)}\right] } \\ \geqslant & \left(\sqrt{\frac{a b}{c}}+\sqrt{\frac{b c}{a}}+\sqrt{\frac{c a}{b}}\right)^{2} \end{aligned}$$

We need to prove that
$$\left(\sqrt{\frac{a b}{c}}+\sqrt{\frac{b c}{a}}+\sqrt{\frac{c a}{b}}\right)^{2} \geqslant 3(a+b+c)$$

Let $\sqrt{\frac{a b}{c}}=x, \sqrt{\frac{b c}{a}}=y, \sqrt{\frac{c a}{b}}=z$, then $a=z x, b=x y, c=y z$, and (2) is equivalent to
$$(x+y+z)^{2} \geqslant 3(x y+y z+z x)$$

This is obvious, so $\frac{a b}{c(c+a)}+\frac{b c}{b(a+b)}+\frac{c a}{c(b+c)} \geqslant \frac{3}{2}$. Therefore, the original inequality is proved.