Từ x+y+z=3 ta có:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\frac{\Leftrightarrow xy+yz+zx}{xyz}=\frac{1}{x+y+z}\)
Nhân chéo ta có:
\(\left(xy+yz+zx\right)\left(x+y+z\right)=xyz\)
\(\Leftrightarrow x^2y+xyz+x^2z+y^2x+y^2z+xyz+xyz+z^2y+z^2x=xyz\)
\(\Leftrightarrow x^2y+x^2z+y^2z+y^2x+z^2x+z^2y+2xyz=0\)
\(\Leftrightarrow\left(x^2y+x^2z+y^2x+xyz\right)+\left(y^2z+z^2x+z^2y+xyz\right)=0\)
\(\Leftrightarrow x\left(xy+xz+y^2+yz\right)+z\left(xy+xz+y^2+yz\right)=0\)
\(\Leftrightarrow\left(x+z\right)\left(xy+xz+y^2+yz\right)=0\)
\(\Leftrightarrow\left(x+z\right)\left[\left(xy+y^2\right)+\left(xz+yz\right)\right]=0\)
\(\Leftrightarrow\left(x+z\right)\left[y\left(x+y\right)+z\left(x+y\right)\right]=0\)
\(\Leftrightarrow\left(x+z\right)\left(y+z\right)\left(x+y\right)=0\)
Suy ra x+z=0 hoặc y+z=0 hoặc x+y=0
Với x+z=0 ta đc y=3
Với y+z=0 ta đc x=3
Với x+y=0 ta đc z=3
Từ đó suy ra đccm
