Proof: Let $x=a+b, y=b+c, z=c+a$, then the original inequality is equivalent to
$$\frac{x(x-y+z)}{y(z+x)}+\frac{y(y-z+x)}{z(x+y)}+\frac{z(z-x+y)}{x(y+z)} \geqslant \frac{3}{2}$$

Notice that
$$\sum \frac{x(x-y+z)}{y(z+x)}=\sum\left(\frac{x}{y}-\frac{z+x}{z+y}\right)+\sum \frac{x}{y+z}$$

By Cauchy's inequality, we have
$$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac{3}{2}$$

From the previous problem, we know
$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \geqslant \frac{z+x}{z+y}+\frac{x+y}{x+z}+\frac{y+z}{y+x}$$

Adding the above two inequalities yields the proof.
Proof: When $a, b, c$ are positive real numbers, we can consider the inverse Ravi substitution to simplify the problem.