Let $a$, $b$ and $c$ be real numbers such that $abc(a+b)(b+c)(c+a)\neq0$ and $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1007}{1008}$ Prove that $\frac{ab}{(a+c)(b+c)}+\frac{bc}{(b+a)(c+a)}+\frac{ca}{(c+b)(a+b)}=2017$
Problem
Source: Regional Olympiad - Federation of Bosnia and Herzegovina 2017
Tags: identity, real numbers, algebra