Prove that since $abc=1$, at least one of $a, b, c$ is not less than 1. Without loss of generality, assume $a \geqslant 1$. Since
$$\begin{aligned} (a+b)(b+c)(c+a)= & (b+c)\left(a^{2}+a b+b c+c a\right) \geq \\ & (b+c)\left(a^{2}+3 \sqrt[3]{(a b c)^{2}}\right)= \\ & (b+c)\left(a^{2}+3\right) \end{aligned}$$

It suffices to prove that
$$\begin{array}{l} (b+c)\left(a^{2}+3\right) \geqslant 4(a+b+c-1) \Leftrightarrow \\ (b+c)\left(a^{2}-1\right) \geqslant 4(a-1) \Leftrightarrow \\ (a-1)((b+c)(a+1)-4) \geqslant 0 \end{array}$$

Given $a \geqslant 1$, it suffices to prove
$$(b+c)(a+1)-4 \geqslant 0$$

And
$$(b+c)(a+1)=a b+c a+b+c=\left(\frac{1}{b}+b\right)+\left(\frac{1}{c}+c\right) \geqslant 4$$

Therefore, the original inequality holds.