Question

設(shè)x,y,z∈(0,1)且x+y+z=2,求W=xy+yz+zx的取值范圍.

Answer 1

解：∵x,y,z,∈(0,1)且x+y+z=2,

∴(1-x)(1-y)(1-z)＞0.

∴1-(x+y+z)+(xy+yz+zx)-xyz＞0.

∴xy+yz+zx＞(x+y+z)-1+xyz=1+xyz＞1.

又x+y+z=2,

∴(x+y+z)²=4.

∴4=x²+y²+z²+2(xy+yz+zx)≥3(xy+yz+zx).

∴3(xy+yz+zx)≤4,即xy+yz+zx≤.

故1＜W≤.