We have :

\(x+y+z=by+cz+ax+cz+ax+by=2\left(ax+by+cz\right)\)

\(=2\left(z+cz\right)=2z\left(1+c\right)\)

or \(=2\left(y+by\right)=2y\left(1+b\right)\)

or \(=2\left(x+ax\right)=2x\left(1+a\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{1+c}=\dfrac{2z}{x+y+z}\\\dfrac{1}{1+b}=\dfrac{2y}{x+y+z}\\\dfrac{1}{1+a}=\dfrac{2x}{x+y+z}\end{matrix}\right.\)\(\Rightarrow\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=\dfrac{2x+2y+2z}{x+y+z}=2\)

P/S : The question must be corrected as shown :

x = by + cz ; y = ax + cz ; z = ax + by