$\frac{x^2}{by+cz}=\frac{y^2}{cz+ax}=\frac{z^2}{ax+by}=2$

Question

If $$\frac{x^2}{by+cz}=\frac{y^2}{cz+ax}=\frac{z^2}{ax+by}=2$$ then find the value of $$\frac{c}{2c+z}+\frac{b}{2b+y}+\frac{a}{2a+x}.$$

I think all the terms need to be manipulated in some way to get the corresponding terms from the expression whose value needs to be found. For example we have to go from $\frac{x^2}{by+cz}$ to $\frac{a}{2a+x}$ in some way or maybe from $\frac{x^2}{by+cz}$ to $\frac{c}{2c+z}$ or something like that. It's very hard to tell from which term to which term I need to go because all the variables are used. So I just tried to make a system of equations.

$$x^2=2(by+cz)$$ $$y^2=2(cz+ax)$$ $$z^2=2(ax+by)$$ $$x^2+y^2+z^2=4(ax+by+cz)$$ $$(x-a)^2+(y-b)^2+(z-c)^2=a^2+b^2+c^2+2(ax+by+cz)$$ But I don't know how to proceed from here.

Answer 1

Hint:

$$x(x+2a)=x^2+2ax=2(ax+by+cz)$$

$$\dfrac1{2a+x}=?$$

$$\dfrac a{2a+x}=\dfrac{ax}{2(ax+by+cz)}$$