Inequality proof $\prod_{cyc} (ab+1)$ Prove that if for positive reals $a,b,c$ with $a^2+b^2+c^2+ab+bc+ca \le 2$ and $a+b+c=1$ then
$$(ab+1)(bc+1)(ca+1)\ge ((1-a)(1-b)(1-c))^2.$$

I've tried expanding and i've noticed that $a+b+c=1$ and $a^2+b^2+c^2+ab+bc+ca\le 4$ imply $a^2+b^2+c^2 \le 3$ but i'm not sure how that helps...

You don't need the additional constraint $a^2+b^2+c^2+ab+bc+ac\leq 2$.
Expanding on my comment, $a+b+c =1 \land a,b,c\geq 0 \implies 0\leq a,b,c \leq 1$. This also implies that
$$0 \leq 1-a, 1-b, 1-c \leq 1$$
Which further implies that
$$(1-a)(1-b)(1-c) \leq 1$$
Which again implies that
$$((1-a)(1-b)(1-c))^2 \leq 1 \quad (*)$$
However,
$$(1+ab)(1+bc)(1+ac) \geq 1\times 1 \times 1 =1 \quad (**)$$
Combining $*, **$, we get
$$(1+ab)(1+bc)(1+ac) \geq 1 \geq  ((1-a)(1-b)(1-c))^2$$
