At first we notice that the equality
$$\begin{aligned} (a+1)(b+1) + (a+1)(c+1) + (b+1)(c+1) &= a+b+c+(a+b+c+2)+ab+ac+bc+1 \\ &= a+b+c+abc+ab+ac+bc+1 = (a+1)(b+1)(c+1) \end{aligned}$$
holds. Now from the inequality between the arithmetic and geometric mean we get:
$$\begin{aligned} \frac{a}{b+1} + \frac{b}{c+1} + \frac{c}{a+1} &= \frac{a+1}{b+1} + \frac{b+1}{c+1} + \frac{c+1}{a+1} - \left(\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}\right) \\ &\ge 3 \\ &\ge 3\sqrt{\frac{(a+1)(b+1)(c+1)}{(b+1)(c+1)(a+1)}} - \left(\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}\right) = 3 \\ &= 3 - \frac{(a+1)(b+1)(c+1)}{(a+1)(b+1)(c+1)} = 3-1=2 \end{aligned}$$
Equality holds if and only if $a=b=c=2$.

If we multiply the inequality by $(a+1)(b+1)(c+1)$ we get the equivalent inequality:
$$a(a+1)(c+1)+b(b+1)(a+1)+c(c+1)(b+1) \geq 2(a+1)(b+1)(c+1)$$
which can be written in the form
$$a^2c+b^2a+c^2b+a^2+b^2+c^2 \geq 2abc+ab+bc+ca+a+b+c+2.$$
But, from the inequalities
$$a^2c+b^2a+c^2b \geq 3\sqrt[3]{a^3b^3c^3} = 3abc$$
$$a^2+b^2+c^2 \geq ab+bc+ca$$
we get
$$a^2c+b^2a+c^2b+a^2+b^2+c^2 \geq 3abc+ab+bc+ca.$$
Now, if we use the condition of the exercise, we get
$$a^2c+b^2a+c^2b+a^2+b^2+c^2 \geq 2abc+ab+bc+ca+a+b+c+2.$$