若实数x,y,z满足x^2+y^2+z^2=1,则xy+yz+zx的取值范围是?[-1/2,1]
问题描述：
若实数x,y,z满足x^2+y^2+z^2=1,则xy+yz+zx的取值范围是?
[-1/2,1]
答
设xy+yz+zx=t
2[x^2+y^2+z^2-(xy+yz+zx)]=(x-y)^2+(x-z)^2+(y-z)^2
2(1-t)=(x-y)^2+(x-z)^2+(y-z)^2>=0
t2[x^2+y^2+z^2+xy+yz+zx]=(x+y)^2+(x+z)^2+(y+z)^2
2(1+t)=(x+y)^2+(x+z)^2+(y+z)^2>=0
t>=-1/2
所以t：[-1/2，1]
答
x^2+y^2+z^2-ab-ac-bc=1/2[(a-c)^2+(b-c)^2+(a-b)^2]
>=0
则1-(ab+bc+ac)>=0
ab+bc+ac=0
则 1+2(ab+bc+ac)>=0
ab+bc+ac