Given $1+a+b+c = 2abc$ and positivity of real numbers $a,b,c$, we are asked to prove that $$\frac{ab}{1+a+b}+\frac{bc}{1+b+c}+\frac{ac}{1+a+c}\geq \frac{3}{2}$$
If $d=a+b+c$ I got as far as to simplify the inequality into $$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d-a}+\dfrac{1}{d-b}+\dfrac{1}{d-c}\geq 3$$ From $a=\frac{1+b+c}{2bc-1}$ I also can prove that $$ab+ac+bc\geq \frac{3}{2}$$ But cannot manage to get to the desired result.
We need to prove next equivalent inequality $$\sum\limits_{cyc}\left(\frac{ab}{1+a+b}+1\right)\geq\frac{9}{2}$$ or $$\sum\limits_{cyc}\frac{(a+1)(b+1)(c+1)}{(1+a+b)(1+c)}\geq\frac{9}{2}$$ Then by Cauchy-Schwarz inequality $$\sum\limits_{cyc}\frac{(a+1)(b+1)(c+1)}{(1+a+b)(1+c)}\geq\frac{9\prod\limits_{cyc}(a+1)}{\sum\limits_{cyc}(1+a+b)(c+1)}=$$ $$=\frac{9\prod\limits_{cyc}(a+1)}{\sum\limits_{cyc}(a+2ab+1+2a)}=\frac{9}{2}$$ Now, we have done!