已知xyz≠0,x≠y,如果(x^2-yz)/[(x(1-yz)]=(y^2-xz)/[y(1-xz)]成立,求证：x+y+z=1/x+1/y+1/z.

证明：(x-(yz/x))/(1-yz)=(y-(xz/y))/(1-xz),十字相乘得：(x-(yz/x))×(1-xz)=（y-(xz/y))×(1-yz),化简：x-(yz/x)-x²z+yz²=y-(xz/y)-y²z+xz²,移项：y-x+yz/x-xz/y+x²z-yz²-y²z+...