x,y,z属于R,且xyz(x+y+z)=1,求证(x+y)(y+z)≥2

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x,y,z属于R,且xyz(x+y+z)=1,求证(x+y)(y+z)≥2

x,y,z属于R,且xyz(x+y+z)=1,求证(x+y)(y+z)≥2
x,y,z属于R,且xyz(x+y+z)=1,求证(x+y)(y+z)≥2

x,y,z属于R,且xyz(x+y+z)=1,求证(x+y)(y+z)≥2
(x+y)(z+y)=xz+y(x+y+z)
因xyz(x+y+z)=1
x+y+z=1/xyz
(x+y)(z+y)
=xz+y(x+y+z)
=xz+1/xz
=(√xy-1/√xy)^2+2>=2

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