Ta có \(x+y+z=2\Leftrightarrow\frac{1}{x+y+z}=\frac{1}{2}\)
Mà \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2}\)
Suy ra \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{xz+yz+z^2}\right)=0\Leftrightarrow\left(x+y\right)\left[\frac{xy+xz+yz+z^2}{xy\left(xz+yz+z^2\right)}\right]=0\Leftrightarrow\frac{\left(x+y\right)\left(x+z\right)\left(y+z\right)}{xy\left(xz+yz+z^2\right)}=0\Leftrightarrow\left(x+y\right)\left(x+z\right)\left(y+z\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
TH1: x=-y\(\Leftrightarrow x+y+z=2\Leftrightarrow\left(-y\right)+y+z=2\Leftrightarrow z=2\)
TH2: y=-z\(\Leftrightarrow x+y+z=2\Leftrightarrow x+\left(-z\right)+z=2\Leftrightarrow x=2\)
TH3: z=-x\(\Leftrightarrow x+y+z=2\Leftrightarrow x+y+\left(-x\right)=2\Leftrightarrow y=2\)
Suy ra có ít nhất một trong ba số x,y,z bằng 2
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+y+z}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Leftrightarrow\frac{y+z}{x\left(x+y+z\right)}+\frac{y+z}{yz}=0\)
\(\Leftrightarrow\left(y+z\right)\left(\frac{1}{x\left(x+y+z\right)}+\frac{1}{yz}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y+z=0\\\frac{1}{x\left(x+y+z\right)}+\frac{1}{yz}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x^2+xy+xz=-yz\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^2+xy+xz+yz=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\left(x+y\right)\left(x+z\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\\left(2-z\right)\left(2-y\right)=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\y=2\\z=2\end{matrix}\right.\)
