$\frac{1}{a+b}+\frac{1}{a+c}=\frac{3}{a+b+c}$ in a triangle Find the angle $\alpha$ of a triangle with sides $a,b$ and $c$ for which the equality $$\dfrac{1}{a+b}+\dfrac{1}{a+c}=\dfrac{3}{a+b+c}$$ holds.
My idea is to use the law of cosines: $$\cos\alpha=\dfrac{b^2+c^2-a^2}{2bc}$$ after simplifying the given equality and plug in something, but this does not seem to be easy.
$$\dfrac{1}{a+b}+\dfrac{1}{a+c}=\dfrac{3}{a+b+c}\\\dfrac{2a+b+c}{(a+b)(a+c)}=\dfrac{3}{a+b+c}\\(2a+b+c)(a+b+c)=3(a+b)(a+c)$$ Am I missing something?

You are doing great, expand the last line, you get $b^2+c^2=a^2+bc$, so the $\cos \alpha = 1/2$.
