Proving $sum_{cyc}frac{(a-1)(c+1)}{1+bc+c}geq 0$ for positive $a$, $b$, $c$ with $abc=1$.

I recently saw the following inequality,
$$frac{(a-1)(c+1)}{1+bc+c}+frac{(b-1)(a+1)}{1+ca+a} + frac{(c-1)(b+1)}{1+ab+b} geq 0 tag1$$
for all $abc=1$ and $a,b,c in mathbb{R}_+ setminus {0 }$.

To prove this inequality, I thought to try to look at each term, by knowing that $abc=1$, and as such we could express the three terms of our inequalities in terms of three/two variables and eliminate the denominator.

First, we shall look at the expression:
$$frac{(a-1)(c+1)}{1+bc+c} = frac{(a-1)(frac{1}{ab}+1)}{1+ b frac{1}{ab}+frac{1}{ab}} = frac{frac{(a-1)(ab+1)}{ab}}{1+ frac{1}{a}+frac{1}{ab}} = frac{(a-1)(ab+1)}{ab(1+ frac{1}{a}+frac{1}{ab})} = frac{(a-1)(ab+1)}{ab+ {b}+{1}}$$
One can see that the denominator now is the same as the denominator of the third term of our inequality. Thus, we shall also bring the denominator of the second term to the form $ab+b+1$. We have

$$frac{(b-1)(a+1)}{1+ca+a}=frac{(b-1)(a+1)}{1+frac{1}{ab}a+a} = frac{(b-1)(a+1)}{1+frac{1}{b}+a} = frac{(b-1)(a+1)b}{b(1+frac{1}{b}+a)} = frac{(b-1)(ab+b)}{b+1+ab} = frac{(b-1)(ab+b)}{ab+b+1}$$

Now we will sum our terms now with the same denominator,
begin{align*}
I &= frac{(a-1)(ab+1)}{ab+ {b}+{1}} + frac{(b-1)(ab+b)}{ab+b+1} +frac{(c-1)(b+1)}{1+ab+b} \ &= frac{(a-1)(ab+1)+(b-1)(ab+b)+(c-1)(b+1)}{ab+ {b}+{1}}
end{align*}
with
$$I=sum_{cyc}frac{(a-1)(c+1)}{1+bc+c} $$
Thus
$$I = frac{a^2b+a-ab-1+ab^2+b^2-ab-b+bc+c-b-1}{ab+ {b}+{1}}$$
So
$$I = frac{a^2b+ab^2 -2(ab+b+1)+b^2+bc+a+c}{ab+ {b}+{1}}$$
Now we must prove that
$$frac{a^2b+ab^2 -2(ab+b+1)+b^2+bc+a+c}{ab+ {b}+{1}} geq 0$$
Since $a,b,c$ are positive real numbers, $ab+b+1 in mathbb{R}_+^*$, so we now are left to prove
$$a^2b+ab^2 -2(ab+b+1)+b^2+bc+a+c geq 0$$
$$a^2b+ab^2+b^2+bc+a+c geq 2(ab+b+1)$$
Note. Here, AM-GM would be a good idea to use. I am currently thinking about a way to use it.
If anybody has a solution or a hint, it’d be much appreciated.