已知xyz=1,x+y+z=2,x²+y²+z²=3,求1/(xy+z-1)+1/(yz+x-1)+1/(zx+y-1)的值

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已知xyz=1,x+y+z=2,x²+y²+z²=3,求1/(xy+z-1)+1/(yz+x-1)+1/(zx+y-1)的值
已知xyz=1,x+y+z=2,x²+y²+z²=3,求1/(xy+z-1)+1/(yz+x-1)+1/(zx+y-1)的值

已知xyz=1,x+y+z=2,x²+y²+z²=3,求1/(xy+z-1)+1/(yz+x-1)+1/(zx+y-1)的值
(x+y+z)^2-(x^2+y^2+z*2)=2xy+2yz+2zx=2^2-3=1
xy+yz+zx=1/2
(x-1)(y-1)(z-1)=xyz-xy-yz-zx+x+y+z-1=1-1/2+2-1=3/2
1/(xy+z-1)+1/(yz+x-1)+1/(zx+y-1)
=1/(xy+1-x-y)+1/(yz+1-y-z)+1/(zx+1-z-x)
=1/[(x-1)(y-1)]+1/[(y-1)(z-1)]+1/[(z-1)(x-1)]
=(z-1+x-1+y-1)/[(x-1)(y-1)(z-1)]
=(2-3)/(3/2)=-2/3

xy=1/z
yz=1/x
zx=1/y
所以f(x)=1/(1/z+z-1)+1/(1/x+x-1)+1/(1/y+y-1)
1/z+z>=2当且仅当z=1等号成立
1/x+x>=2当且仅当x=1等号成立
1/y+y>=2当且仅当y=1等号成立
此时x=y=1
f(x)=1+1+1=3

(x+y+z)^2-(x^2+y^2+z^2)=2xy+2yz+2zx=2^2-3=1
xy+yz+zx=1/2
(x-1)(y-1)(z-1)=xyz-xy-yz-zx+x+y+z-1=1-1/2+2-1=3/2
原式=1/(xy+1-x-y)+1/(yz+1-y-z)+1/(zx+1-z-x)
=1/[(x-1)(y-1)]+1/[(y-1)(z-1)]+1/[(z-1)(x-1)]
=(z-1+x-1+y-1)/[(x-1)(y-1)(z-1)]
=(x+y+z-3)/[(x-1)(y-1)(z-1)]
=(2-3)/(3/2)=-2/3

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