Solution:
The required inequality is equivalent to
$$\frac{c(a+1)+a(b+1)+b(c+1)}{a b c(a+1)(b+1)(c+1)} \geq \frac{3}{(1+a b c)^{2}}$$
or equivalently to,
$$(1+a b c)^{2}(a b+b c+c a+a+b+c) \geq 3 a b c(a b+b c+c a+a+b+c+a b c+1)$$
Let $m=a+b+c, n=a b+b c+c a$ and $x^{3}=a b c$, then the above can be rewritten as
$$(m+n)\left(1+x^{3}\right)^{2} \geq 3 x^{3}\left(x^{3}+m+n+1\right)$$
or
$$(m+n)\left(x^{6}-x^{3}+1\right) \geq 3 x^{3}\left(x^{3}+1\right)$$
By the AM-GM inequality we have $m \geq 3 x$ and $n \geq 3 x^{2}$, hence $m+n \geq 3 x(x+1)$. It is sufficient to prove that
$$\begin{aligned} x(x+1)\left(x^{6}-x^{3}+1\right) & \geq x^{3}(x+1)\left(x^{2}-x+1\right) \\ 3\left(x^{6}-x^{3}+1\right) & \geq x^{2}\left(x^{2}-x+1\right) \\ \left(x^{2}-1\right)^{2} & \geq 0 \end{aligned}$$
which is true.

Alternative solution by PSC. We present here an approach without fully expanding.
Let $a b c=k^{3}$ and set $a=k \frac{x}{y}, b=k \frac{y}{z}, c=k \frac{z}{x}$, where $k, x, y, z>0$. Then, the inequality can be rewritten as
$$\frac{z^{2}}{(k y+z)(k z+x)}+\frac{x^{2}}{(k z+x)(k x+y)}+\frac{y^{2}}{(k x+y)(k y+z)} \geq \frac{3 k^{2}}{\left(1+k^{3}\right)^{2}}$$
Using the Cauchy-Schwarz inequality we have that
$$\sum_{\text{cyclic}} \frac{z^{2}}{(k y+z)(k z+x)} \geq \frac{(x+y+z)^{2}}{(k y+z)(k z+x)+(k z+x)(k x+y)+(k x+y)(k y+z)}$$
therefore it suffices to prove that
$$\frac{(x+y+z)^{2}}{(k y+z)(k z+x)+(k z+x)(k x+y)+(k x+y)(k y+z)} \geq \frac{3 k^{2}}{\left(1+k^{3}\right)^{2}}$$
or
$$\left(\left(1+k^{3}\right)^{2}-3 k^{3}\right)\left(x^{2}+y^{2}+z^{2}\right) \geq\left(3 k^{2}\left(k^{2}+k+1\right)-2\left(1+k^{3}\right)^{2}\right)(x y+y z+z x)$$
Since $x^{2}+y^{2}+z^{2} \geq x y+y z+z x$ and $\left(1+k^{3}\right)^{2}-3 k^{3}>0$, it is enough to prove that
$$\left(1+k^{3}\right)^{2}-3 k^{3} \geq 3 k^{2}\left(k^{2}+k+1\right)-2\left(1+k^{3}\right)^{2}$$
or
$$(k-1)^{2}\left(k^{2}+1\right)(k+1)^{2} \geq 0$$
which is true.