Ta có:
$\left\{\begin{array}{l}x=by+cz\\y=cz+ax\\z=ax+by\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}ax=\dfrac{y+z-x}{2}\\by=\dfrac{x+z-y}{2}\\cz=\dfrac{x+y-z}{2}\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x(1+a)=\dfrac{y+z+x}{2}\\y(1+b)=\dfrac{x+z+y}{2}\\z(1+c)=\dfrac{x+y+z}{2}\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}\dfrac{1}{1+a}=\dfrac{2x}{x+y+z}\\\dfrac{1}{1+b}=\dfrac{2y}{x+y+z}\\\dfrac{1}{1+c}=\dfrac{2z}{x+y+z}\end{array}\right.$
$\Rightarrow \dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=2$