The inequality is equivalent to
$$\frac{c(a+1)+a(b+1)+b(c+1)}{abc(a+1)(b+1)(c+1)} \ge \frac{3}{(1+abc)^2},$$
or after simplifications to
$$(1+abc)^2(ab+bc+ca+a+b+c) \ge 3abc(ab+bc+ca+a+b+c+abc+1).$$
We put $m = a+b+c$, $n = ab+bc+ca$ and $x^3 = abc$, to obtain:
$$(m+n)(1+x^3)^2 \ge 3x^3(x^3 + m+n+1),$$
or $(m+n)(x^6 - x^3 + 1) \ge 3x^3(x^3 + 1).$
From the inequality of arithmetic–geometric mean we have $m \ge 3x$ and $n \ge 3x^2$, and so $m+n \ge 3x(x+1)$. Therefore, it is enough to prove that
$$\begin{aligned} & 3x(x+1)(x^6-x^3+1) \ge 3x^3(x+1)(x^2-x+1) \\ \Leftrightarrow & x^6-x^3+1 \ge x^2(x^2-x+1) \\ \Leftrightarrow & x^6-x^4-x^2+1 \ge 0 \Leftrightarrow x^4(x^2-1)-(x^2-1) \ge 0 \\ \Leftrightarrow & (x^2-1)(x^4-1) \ge 0 \Leftrightarrow (x^2+1)(x^2-1)^2 \ge 0, \end{aligned}$$
which is valid.