Given $\triangle ABC$ with $C=60^\circ$, show that $\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$ 
Given that $C=60^\circ$ on a triangle $ABC$, prove the following relation: $$\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$$

P.S. Maybe this info could be of help: I used the cosine rule of triangles, given that $C=60°$. 

Let's try to simplify the expression $\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$.
Multiply through by $a + b + c$ and you obtain the equivalent expression
$$\frac{a+b+c}{a+c} + \frac{a + b + c}{b + c} = 3$$
This is equivalent to
$$\frac{b}{a+c} + \frac{a}{b + c} = 1$$
Multiplying through by $(a+c)(b+c)$ this is the same as
$$b(b+c) + a(a + c) = (a + c)(b+c)$$
Expanding out this becomes
$$b^2 + bc + a^2 + ac = ab + cb + ac + c^2$$
Cancelling terms this becomes
$$b^2 + a^2 - ab = c^2$$
This is exactly what the law of cosines gives you.
