Prove $\frac{ab}{1+a+b}+\frac{bc}{1+b+c}+\frac{ac}{1+a+c}\geq \frac{3}{2}$ for $a$, $b$, $c$ positive and $1+a+b+c=2abc$ 
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!
