问题: 题3
解答:
解：
(1)
`(x+y)(y+z)
=xz+yz+xy+y²
=xz+y(x+y+z)
≥2√[xz*y(x+y+z)]
=2
当xz=y(x+y+z)=1/xz,即xz=1时取得最小值2.
(2)
xyz(x+y+z)=1 => x+y+z=1/(xyz)
`(x+y)(y+z)
=[1/(xyz)-z][1/(xyz)-x)]
=[1/(xyz)]²-[1/(yz)+1/(xy)]+xz
=(x+y+z)[1/(xyz)]-[1/(yz)+1/(xy)]+xz
=1/(yz)+1/(xy)+1/(xz)-[1/yz-1/xy]+xz
=1/(xz)+(xz)
≥2
xz=1时,(x+y)(y+z)的最小值为2
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