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

If $d=a+b+c$ I got as far as to simplify the inequality into

$\frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}}+{\displaystyle \frac{1}{d-a}}+{\displaystyle \frac{1}{d-b}}+{\displaystyle \frac{1}{d-c}}\ge 3$

From $a=\frac{1+b+c}{2bc-1}$ I also can prove that

$ab+ac+bc\ge \frac{3}{2}$

But cannot manage to get to the desired result.

If $d=a+b+c$ I got as far as to simplify the inequality into

$\frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}}+{\displaystyle \frac{1}{d-a}}+{\displaystyle \frac{1}{d-b}}+{\displaystyle \frac{1}{d-c}}\ge 3$

From $a=\frac{1+b+c}{2bc-1}$ I also can prove that

$ab+ac+bc\ge \frac{3}{2}$

But cannot manage to get to the desired result.