Proof. For convenience, we write the original inequality as
$$(x+y)(y+z)(z+x) \geqslant 4(x+y+z-1)$$

where $xyz=1$. Making the substitution $x=s-a, y=s-b, z=s-c$, then $a, b, c$ are the side lengths of a triangle, and $s$ is the semi-perimeter of the triangle. Thus, the inequality becomes
$$abc \geqslant 4(s-1)$$

Since
$$xyz=1 \Leftrightarrow s(s-a)(s-b)(s-c)=s \Leftrightarrow s r^{2}=1$$

we only need to prove
$$\begin{array}{c} 4 R r s \geqslant 4(s-1) \Leftrightarrow R \geqslant(s-1) r \Leftrightarrow R+r \geqslant s r \Leftrightarrow \\ \quad(R+r)^{3} \geqslant s^{3} r^{3}=s^{2} r \end{array}$$

By the AM-GM inequality, we have
$$R+r=\frac{R}{2}+\frac{R}{2}+r \geqslant 3 \sqrt[3]{\frac{R^{2} r}{4}}$$

We only need to prove
$$s \leqslant \frac{3 \sqrt{3} R}{2}$$