Proof: Let $\sqrt[3]{a b c}=k(k>0)$, then $a b c=k^{3}$, so we can set $a=k \frac{a_{2}}{a_{1}}, b=k \frac{a_{3}}{a_{2}}, c=k \frac{a_{1}}{a_{3}}$, $\left(a_{1}, a_{2}, a_{3}>0\right)$, substituting into (1), we only need to prove
$$\frac{1}{k \frac{a_{2}}{a_{1}}+k^{2} \frac{a_{3}}{a_{1}}}+\frac{1}{k \frac{a_{3}}{a_{2}}+k^{2} \frac{a_{1}}{a_{2}}}+\frac{1}{k \frac{a_{1}}{a_{3}}+k^{2} \frac{a_{2}}{a_{3}}} \geqslant \frac{3}{k(1+k)}$$

which is
$$\frac{a_{1}}{a_{2}+k a_{3}}+\frac{a_{2}}{a_{3}+k a_{1}}+\frac{a_{3}}{a_{1}+k a_{2}} \geqslant \frac{3}{1+k}$$

Below we prove inequality (2) (using the rearrangement inequality).
Proof 1: Let $x=a_{2}+k a_{3}, y=a_{3}+k a_{1}, z=a_{\llcorner}+k a_{2}$, then $x, y, z$ are all positive numbers, solving we get.
$$a_{1}=\frac{-k x+k^{2} y+z}{1+k^{3}} a_{2}=\frac{-k y+k^{2} z+x}{1+k^{3}} a_{3}=\frac{-k z+k^{2} x+y}{1+k^{3}}$$

So the left side of inequality (2) can be transformed into
$$\frac{-3 k}{1+k^{3}}+\frac{k^{2}}{1+k^{3}}\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right)+\frac{1}{1+k^{3}}\left(\frac{z}{x}+\frac{x}{y}+\frac{y}{z}\right)$$

By the AM-GM inequality, $\frac{y}{x}+\frac{z}{y}+\frac{x}{z} \geqslant 3, \frac{z}{x}+\frac{x}{y}+\frac{y}{z} \geqslant 3$, so
$$\frac{a_{1}}{a_{2}+k a_{3}}+\frac{a_{2}}{a_{3}+k a_{1}}+\frac{a_{3}}{a_{1}+k a_{2}} \geqslant \frac{-3 k}{1+k^{3}}+\frac{3 k^{2}}{1+k^{3}}+\frac{3}{1+k^{3}}=\frac{3}{1+k}$$

Proof 2: By the Cauchy-Schwarz inequality, we have
$$\left(\frac{x_{1}^{2}}{y_{1}}+\frac{x_{2}^{2}}{y_{2}}+\frac{x_{3}^{2}}{y_{3}}\right)\left(y_{1}+y_{2}+y_{3}\right) \geqslant\left(x_{1}+x_{2}+x_{3}\right)^{2}$$

which is
$$\frac{x_{1}^{2}}{y_{1}}+\frac{x_{2}^{2}}{y_{2}}+\frac{x_{3}^{2}}{y_{3}} \geqslant \frac{\left(x_{1}+x_{2}+x_{3}\right)^{2}}{y_{1}+y_{2}+y_{3}}$$
(Where $y_{1}, y_{2}, y_{3}$ are positive numbers, and $x_{1}, x_{2}, x_{3}$ are any real numbers)
Thus,
$$\begin{array}{l} \frac{a_{1}}{a_{2}+k a_{3}}+\frac{a_{2}}{a_{3}+k a_{1}}+\frac{a_{3}}{a_{1}+k a_{2}}= \\ \frac{a_{1}^{2}}{a_{1}\left(a_{2}+k a_{3}\right)}+\frac{a_{2}^{2}}{a_{2}\left(a_{3}+k a_{1}\right)}+\frac{a_{3}^{2}}{a_{3}\left(a_{1}+k a_{2}\right)} \geqslant \\ \frac{\left(a_{1}+a_{2}+a_{3}\right)^{2}}{(1+k)\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}\right)} \geqslant \frac{3}{1+k} \end{array}$$

where the last step uses the inequality $\left(a_{1}+a_{2}+a_{3}\right)^{2} \geqslant 3\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}\right)$, which is obvious.

Proof 3:
$$\begin{array}{l} (1+a b c)\left(\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\right)+3= \\ \left(\frac{1+a b c}{a+a b}+1\right)+\left(\frac{1+a b c}{b+b c}+1\right)+\left(\frac{1+a b c}{c+c a}+1\right)= \\ \frac{(a b c+a b)+(1+a)}{a+a b}+\frac{(a b c+b c)+(1+b)}{b+b c}+\frac{(a b c+c a)+(1+c)}{c+c a}= \\ \frac{a+1}{a(1+b)}+\frac{b+1}{b(1+c)}+\frac{c+1}{c(1+a)}+\frac{b(c+1)}{1+b}+\frac{c(a+1)}{1+c}+\frac{a(b+1)}{1+a} \geqslant \\ 3\left(\sqrt[3]{\frac{a+1}{a(1+b)} \cdot \frac{b+1}{b(1+c)} \cdot \frac{c+1}{c(1+a)}}+\sqrt[3]{\frac{b(c+1)}{1+b} \cdot \frac{c(a+1)}{1+c} \cdot \frac{a(b+1)}{1+a}}\right)= \\ 3\left(\sqrt[3]{a b c}+\frac{1}{\sqrt[3]{a b c}}\right) \end{array}$$

Thus,
$$\begin{array}{l} (1+a b c)\left(\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\right) \geqslant 3\left(\sqrt[3]{a b c}+\frac{1}{\sqrt[3]{a b c}}-1\right) \\ \frac{\sqrt[3]{a b c}+\frac{1}{\sqrt[3]{a b c}}-1}{1+a b c} \geqslant \frac{1}{\sqrt[3]{a b c}(1+\sqrt[3]{a b c})} \text{. This is an identity. } \end{array}$$