证明:由已知 x+
1
y =y+
1
z =z+
1
x 得出:
∵x+
1
y =y+
1
z ,
∴x-y=
1
z -
1
y ,
x-y=
y-z
yz ,
∴yz=
y-z
x-y ,①
同理得出
zx=
z-x
y-z ,②
xy=
x-y
z-x .③
①×②×③得x 2y 2z 2=1.
证明:由已知 x+
1
y =y+
1
z =z+
1
x 得出:
∵x+
1
y =y+
1
z ,
∴x-y=
1
z -
1
y ,
x-y=
y-z
yz ,
∴yz=
y-z
x-y ,①
同理得出
zx=
z-x
y-z ,②
xy=
x-y
z-x .③
①×②×③得x 2y 2z 2=1.