已知:xyz∈R+且x+y+z=1,求证:(1-x)(1-y)(1-z)≥8xyz
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已知:xyz∈R+且x+y+z=1,求证:(1-x)(1-y)(1-z)≥8xyz
该如何证明?
该如何证明?
![已知:xyz∈R+且x+y+z=1,求证:(1-x)(1-y)(1-z)≥8xyz](/uploads/image/z/16692877-37-7.jpg?t=%E5%B7%B2%E7%9F%A5%3Axyz%E2%88%88R%2B%E4%B8%94x%2By%2Bz%3D1%2C%E6%B1%82%E8%AF%81%3A%281-x%29%281-y%29%281-z%29%E2%89%A58xyz)
1-x=y+z
1-y=x+z
1-z=x+y
由题意知x>0,y>0,z>0
y+z>=2根号y*根号z
x+z>=2根号x*根号z
y+x>=2根号y*根号x
(1-x)(1-y)(1-z)>=2根号y*根号z*2根号x*根号z*2根号y*根号x=8xyz
即(1-x)(1-y)(1-z)≥8xyz
1-y=x+z
1-z=x+y
由题意知x>0,y>0,z>0
y+z>=2根号y*根号z
x+z>=2根号x*根号z
y+x>=2根号y*根号x
(1-x)(1-y)(1-z)>=2根号y*根号z*2根号x*根号z*2根号y*根号x=8xyz
即(1-x)(1-y)(1-z)≥8xyz
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