Proof: By the AM-GM inequality, we have
$$L H S \geq 3 \sqrt[3]{\sqrt{\frac{a+b}{a+1}} \cdot \sqrt{\frac{b+c}{b+1}} \cdot \sqrt{\frac{c+a}{c+1}}}=3 \sqrt[6]{\frac{(a+b)(b+c)(c+a)}{(a+1)(b+1)(c+1)}}$$

Thus, we only need to prove
$$(a+b)(b+c)(c+a) \geq(a+1)(b+1)(c+1)$$

Since $a b c=1$, $(*)$ is equivalent to
$$a b(a+b)+b c(b+c)+c a(c+a) \geq a+b+c+a b+b c+c a$$

By the AM-GM inequality, we have
$$\begin{array}{l} 2 L H S+\sum_{c y c} a b=\sum_{c y c}\left(a^{2} b+a^{2} b+a^{2} c+a^{2} c+b c\right) \geq 5 \sum_{c y c} a \\ 2 L H S+\sum_{c y c} a=\sum_{c y c}\left(a^{2} b+a^{2} b+b^{2} a+b^{2} a+c\right) \geq 5 \sum_{c y c} a b \end{array}$$

Therefore,
$$4 L H S+2 \sum_{c y c} a b+\sum_{c y c} a \geq 5 \sum_{c y c} a b+4 \sum_{c y c} a b \Rightarrow 4 L H S \geq 4 \sum_{c y c} a+4 \sum_{c y c} a b=4 R H S$$

Thus, the original inequality holds. Equality holds if and only if $a=b=c=1$.