Question

若实数x,y,z满足x²+y²+z²=1,则xy+yz+zx的取值范围是(    )

A.［-1,1］           B.［-,1］           C.［-1,］           D.［-,］

Answer 1

解析:∵xy+yz+zx≤++=x²+y²+z²=1,

    又∵2(xy+yz+zx)=(x+y+z)²-(x²+y²+z²)≥0-1=-1,

    ∴xy+yz+zx≥-.故选择B.

答案:B