Sửa đề: Không cần giả thiết $x+y+z=0$

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

Lại có :

$\left\{ \begin{array}{l} x = by + cz\\ y = ax + cy\\ z = ax + by \end{array} \right. \Rightarrow x + y + z = 2\left( {ax + by + cz} \right)$

$ \Rightarrow \dfrac{1}{{a + 1}} + \dfrac{1}{{b + 1}} + \dfrac{1}{{c + 1}} = \dfrac{{x + y + z}}{{ax + by + cz}} = \dfrac{{2\left( {ax + by + cz} \right)}}{{ax + by + cz}} = 2$

$\begin{cases} x=by+cz(1)\\y=ax+cz(2)\\z=ax+by(3)\end{cases}\\(1)+(2)+(3)\to x+y+z=2(ax+by+cz)\\\to x+y+z=2(ax+x)\\\to x+y+z=2x(a+1)\\\to a+1=\dfrac{x+y+z}{2x}\\\to \dfrac{1}{a+1}=\dfrac{2x}{x+y+z}$

Tương tự: $\begin{cases} \dfrac{1}{b+1}=\dfrac{2y}{x+y+z}\\\dfrac{1}{c+1}=\dfrac{2z}{x+y+z}\end{cases}$

$\to \dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}=\dfrac{2(x+y+z)}{x+y+z}=2$