We first prove that if $a, b > 0$, then
$$\frac{1}{(1+a)^2} + \frac{1}{(1+b)^2} \ge \frac{1}{1+ab}$$
with equality if and only if $a = b = 1$. To prove this we multiply through by $(1+a)^2(1+b)^2(1+ab)$ and simplify to obtain the equivalent inequality $a^3b+ab^3-a^2b^2-2ab+1 \ge 0$. This can be written as $ab(a-b)^2+(ab-1)^2 \ge 0$, which is clearly true. Equality occurs if and only if $a = b$ and $ab = 1$, that is $a = b = 1$.

Similarly
$$\frac{1}{(1+b)^2} + \frac{1}{(1+c)^2} \ge \frac{1}{1+bc} \quad \text{with equality if and only if } b = c = 1,$$
$$\frac{1}{(1+c)^2} + \frac{1}{(1+a)^2} \ge \frac{1}{1+ca} \quad \text{with equality if and only if } c = a = 1.$$
Adding the three inequalities gives the desired result.