由已知條件：
x+y+z=2
x^2+y^2+z^2=3
所以xy+yz+zx=(1/2)[(x+y+z)^2-(x^2+y^2+z^2)]=1/2
又因為左式第一項
1/(xy+z-1)=1/[xy+(2-x-y)-1]=1/[(x-1)(y-1)]
同理
1/(yz+x-1)=1/[(y-1)(z-1)]
1/(zx+y-1)=1/[(z-1)(x-1)]
三式相加（此時通分便很簡單）得：
(3-x-y-z)/[(1-x)(1-y)(1-z)]
1/[(1-x)(1-y)(1-z)]
=1/(1-x-y-z+xy+yz+zx-xyz)
=1/(1-2+1/2-1)
=-2/3