Question

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

Answer 1

(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