Solution:
We have
$A=(a+b)(b+c)(c+a)=\left(a b+a c+b^{2}+b c\right)(c+a)=(b(a+b+c)+a c)(c+a)$,
so by the given condition
$$A=\left(\frac{3}{a c}+a c\right)(c+a)=\left(\frac{1}{a c}+\frac{1}{a c}+\frac{1}{a c}+a c\right)(c+a)$$
Applying the AM-GM inequality for four and two terms respectively, we get
$$A \geq 4 \sqrt[4]{\frac{a c}{(a c)^{3}}} \cdot 2 \sqrt{a c}=8$$
From the last part, it is easy to see that equality holds when $a=c$ and $\frac{1}{a c}=a c$, i.e. $a=b=c=1$.