that looks hard

gobathegreat wrote: 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 $\frac{ab}{(a+c)(b+c)}+\frac{bc}{(b+a)(c+a)}+\frac{ca}{(c+b)(a+b)}=2017$ $$\frac{ab}{(a+c)(b+c)}+\frac{bc}{(b+a)(c+a)}+\frac{ca}{(c+b)(a+b)}+\frac{2abc}{(a+b)(b+c)(c+a)}=1$$

Hello. Let us factorize this. $\frac{ab}{(b+c)(a+c)}+\frac{bc}{(c+a)(a+b)}+\frac{ca}{(a+b)(b+c)}=\frac{ab(a+b)+bc(b+c)+ca(a+c)}{(a+b)(b+c)(c+a)}=\frac{a^2b+ab^2+b^2c+bc^2+c^2a+ca^2}{2abc+a^2b+ac^2+a^2b+b^2c+b^2a+bc^2}$. We will factorize this again, so: $\frac{(a+b+c)(ab+bc+ca)-3abc}{(a+b+c)(ab+bc+ca)-abc}=\frac{abc((a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-3)}{abc((a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-1)}=\frac{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-3}{(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-1}$. Now we will use given facts, so this is equal to: $\frac{\frac{1007}{1008}-3}{\frac{1007}{1008}-1}=2017$. That is all