Three Fractions - Repovive Math Contest Round 2

Three Fractions - Repovive Math Contest Round 2 | Repovive

Let $a,b,c$ be positive real numbers, none equal to $1$ , such that

a+b+c=ab+bc+ca.

Prove that

\begin{aligned} &\frac{(bc-b-c)(1-a)}{(bc-a)(1-b-c)} +\frac{(ab-a-b)(1-c)}{(ab-c)(1-a-b)}\\ &\qquad+\frac{(ca-a-c)(1-b)}{(ca-b)(1-a-c)} \ge \frac{a+b+c}{2}. \end{aligned}

Three Fractions 1000 pts