Ta có x + y = 2cz + ax + by = 2cz + z
hay 2cz = x + y - z, suy ra c = \(\frac{x+y-z}{2z}\)
do đó: \(1+c=\frac{x+y+z}{2z}\) hay \(\frac{1}{1+c}=\frac{2z}{z+y+z}\)
Tương tự \(1+a=\frac{x+y+z}{2x}\) hay \(\frac{1}{1+a}=\frac{2x}{x+y+z}\)
\(1+b=\frac{x+y+z}{2y}\) hay \(\frac{1}{1+b}=\frac{2y}{x+y+z}\)
Vậy \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
Ta có \(\left\{\begin{matrix}x=by+cz\\y=ax+cz\\z=ax+by\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}ax+x=ax+by+cz\\by+y=ax+by+cz\\cz+z=ax+by+cz\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x\left(a+1\right)=ax+by+cz\\y\left(b+1\right)=ax+by+cz\\z\left(c+1\right)=ax+by+cz\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}a+1=\frac{ax+by+cz}{x}\\b+1=\frac{ax+by+cz}{y}\\c+1=\frac{ax+by+cz}{z}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{1}{a+1}=\frac{x}{ax+by+cz}\\\frac{1}{b+1}=\frac{y}{ax+by+cz}\\\frac{1}{c+1}=\frac{z}{ax+by+cz}\end{matrix}\right.\)
\(\Rightarrow\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{x+y+z}{ax+by+cz}\)
Ta lại có \(\left\{\begin{matrix}x=by+cz\\y=ax+cz\\z=ax+by\end{matrix}\right.\)
\(\Rightarrow x+y+z=2\left(ax+by+cz\right)\)
\(\Rightarrow\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{x+y+z}{ax+by+cz}=\frac{2\left(ax+by+cz\right)}{ax+by+cz}=2\)
Vậy \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2\left(đpcm\right)\)
