证明:|1/x+1/y+1/z|^2=1/x^2+1/y^2+1/z^2+2/(xy)+2/(yz)+2/(zx)
=1/x^2+1/y^2+1/z^2+2/(xy)+2/z(1/x+1/y)
=1/x^2+1/y^2+1/z^2+2/(xy)+2/z[(x+y)/xy]
因为z=-x-y
:|1/x+1/y+1/z|^2=1/x^2+1/y^2+1/z^2+2/(xy)-2/(x+y)[(x+y)/xy]
=1/x^2+1/y^2+1/z^2
两边开根号,得到结论
证毕
若x,y,z均不为0,且x+y+z=0,证明：√（1/x^2+1/y^2+1/z^2)=|1/x+1/y+1/z|
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