证明：∵x²+y²+z²=xy+yz+zx
∴x²+y²+z²-xy-yz-zx=0
两边同时乘以2,得
2x²+2y²+2z²-2xy-2yz-2zx=0
即x²-2xy+y²+y²-2yz+z²+x²-2zx+z²=0
∴(x-y)²+(y-z)²+(x-z)²=0
∵(x-y)²≥0,(y-z)²≥0,(x-z)²≥0
∴(x-y)²=0,(y-z)²=0,(x-z)²=0
∴x-y=0,y-z=0,x-z=0
∴x=y,y=z,x=z
即x=y=z