x,y,z为实数 且(y-z)^2+(x-y)^2+(z-x)^2=(y+z-2x)^2+(x+z-

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(y-z)^2+(z-x)^2+(x-y)^2=(x+y-2z)^2+(y+z-2x)^2+(z+x-2y)^2
[(y-z)^2-(y+z-2x)^2]+[(z-x)^2-(x+z-2y)^2]+[(x-y)^2-(x+y-2z)^2]=0
[-4(y-x)(z+x)]+[-4(z-y)(x+y)]+[-4(x-z)(y+z)]=0
(yz-xz+xy-x^2)+(xz-xy+yz-y^2)+(xy-yz+xz-y^2)=0
xy+xz+yz=x^2+y^2+z^2
(xy+xz+yz)+(x^2y^2z^2+x^2yz+xy^2z+xyz^2+1)=(x^2+y^2+z^2)+
(x^2y^2z^2+x^2yz+xy^2z+xyz^2+1)
(yz+1)(zx+1)(xy+1)=(x^2+1)(y^2+1)(z^2+1)
所以 [(yz+1)(zx+1)(xy+1)]/[(x^2+1)(y^2+1)(z^2+1)]=1
======以下答案可供参考======
供参考答案1:
设a=y-z,b=x-y,c=z-x,显然a+b+c=0
则原式a²+b²+c²=(c-b)²+(b-a)²+(a-c)²
=>a²+b²+c²=2ab+2bc+2ca
(a+b+c)²=0
a²+b²+c²+2ab+2bc+2ca=0
则2(a²+b²+c²)=0
故a=b=c=0
则x=y=z
则(yz+1)(zx+1)(xy+1)\(x^2+1)(y^2+1)(z^2+1) =1