Question

实数x,y,z满足x+y+z=1/x+1/y+1/z=1,求证:x,y,z中至少有一个为1

Answer 1

x+y+z=1 ==> x+y=1-z -----(1)
1/x+1/y+1/z=1 ==>(x+y)/(xy)=1-1/z=(z-1)/z
因为x+y=1-z得:(1-z)/(xy)=(z-1)/z ------------(2)
若z=1,题目得证,
若z<>1,则(2)式两边处以(1-z)得:1/(xy)=-1/z
==>xy=-z ------------(3)
由(1)得:-z=x+y-1, 代入3得:
xy=x+y-1 ==>xy-x-y+1=0 ----------(4)
将(4)分解因式得: (x-1)*(y-1)=0
解得:x=1 或 y=1 问题得证

另外方法:
x+y+z=1 ==> x+y+z-1=0 --------(1)
1/x+1/y+1/z=1 ==> xyz-xy-yz-zx=0 --------(2)
(1)+(2)得:xyz-xy-yz-zx+x+y+z-1=0
将左边进行因式分解得:
xyz-xy-yz-zx+x+y+z-1
=(xyz-xy)+(-yz+z)+(-zx+x)+(z-1)
=xy(z-1)-y(z-1)-x(z-1)+(z-1)
=(z-1)(xy-y-x+1)
=(z-1)(x-1)(y-1)
=0
所以:x-1=0 或 y-1=0 或 z-1=0
即: x=1 或 y=1 或 z=1
问题得证