Solution:
We have $a b+4=\frac{8}{c}+4=\frac{4(c+2)}{c}$ and similarly $b c+4=\frac{4(a+2)}{a}$ and $c a+4=\frac{4(b+2)}{b}$. It follows that
$$(a b+4)(b c+4)(c a+4)=\frac{64}{a b c}(a+2)(b+2)(c+2)=8(a+2)(b+2)(c+2)$$
so that
$$\frac{(a b+4)(b c+4)(c a+4)}{(a+2)(b+2)(c+2)}=8$$
Applying AM-GM, we conclude:
$$\frac{a b+4}{a+2}+\frac{b c+4}{b+2}+\frac{c a+4}{c+2} \geq 3 \cdot \sqrt[3]{\frac{(a b+4)(b c+4)(c a+4)}{(a+2)(b+2)(c+2)}}=6$$
Alternatively, we can write LHS as
$$\frac{b c(a b+4)}{2(b c+4)}+\frac{a c(b c+4)}{2(a c+4)}+\frac{a b(c a+4)}{2(a b+4)}$$
and then apply AM-GM.