$\frac{1}{a+b}+\frac{1}{a+c}=\frac{3}{a+b+c}$ 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?

Multiplying both sides of the equation by $(a+b)(a+c)(a+b+c)$, we get \begin{align*}
(a+c)(a+b+c)+(a+b)(a+b+c)&=3(a+c)(a+b)\\
a^2 + a b + 2 a c + b c + c^2+a^2 + 2 a b + a c + b^2 + b c&=3 a^2 + 3 a b + 3 a c + 3 b c\\
2 a^2 + 3 a b + 3 a c + b^2 + 2 b c + c^2&=3 a^2 + 3 a b + 3 a c + 3 b c\\
b c &=-a^2+ b^2 + c^2\\
\end{align*}
Hence, we have $\cos\alpha=\frac{1}{2}$. Since we know $0<\alpha<\pi$, we have $\alpha=\boxed{\frac{\pi}{3}}$
