Prove $\frac{1}{2a+2bc+1} + \frac{1}{2b+2ca+1} + \frac{1}{2c+2ab+1} \ge 1$

If none of the terms $a+bc$, $b+ca$, and $c+ab$ is greater than one, then each summand of the LHS is at least $\frac{1}{3}$, and thus the inequality holds. It remains to show the inequality also holds when at least one of $a+bc$, $b+ca$, or $c+ab$ is strictly greater than one. WOLOG, suppose that $c+ab>1$. Since $c=\frac{1-ab}{a+b}$, we must have $a+b<1$. Furthermore, from Cauchy-Schwarz inequality we have $$\frac{1}{2a+2bc+1}+\frac{1}{2b+2ca+1}\geq \frac{4}{2a+2bc+1+2b+2ca+1}\\=\frac{2}{2+a+b-ab}.$$ Therefore, it suffices to show that$$\frac{2}{2+a+b-ab}+\frac{1}{2c+2ab+1} = \frac{2}{2+a+b-ab}+\frac{a+b}{2+a+b-2ab+2ab(a+b)}\\ \geq 1, $$ holds for all $a+b<1$. We have $$A:=\frac{2}{2+a+b-ab}+\frac{a+b}{2+a+b-2ab+2ab(a+b)}\\=\frac{4+4(a+b)-4ab+3ab(a+b)+(a+b)^2}{(2+a+b-ab)(2+a+b-2ab+2ab(a+b))}.$$ For compactness we can let $s=a+b$ and $p=ab$ and obtain $$A=\frac{4+4s-4p+3sp+s^2}{(2+s-p)(2+s-2p+2sp)}\\=\frac{4+4s-4p+3sp+s^2}{4+4s-6p+3sp+s^2+2s^2p-2p^2+2sp^2}.$$ Hence, we need to show that $$2p\geq 2s^2p-2p^2+2sp^2,$$ which is equivalent to $$2p(1-s)(1+p+s)\geq 0.$$ This inequality is valid because by assumption $s<1$. Equation cannot hold in the last case, since we need to have $p=0$ which implies $a$ or $b$ is zero, but the Cauchy-Schwarz we applied above cannot hold for these values considering $a+b<1$. The only remaining possibility for having equation is in the first case where we should have $a+bc=b+ca=a+bc=1$ which reduces to $(a,b,c)\in\{(0,1,1),(1,0,1),(1,1,0)\}$.