令2000x^3=2001y^3=2002z^3=N,题中方程立方得
2000x^2+2001y^2+2000z^2=（3次√2000+3次√2001+3次√2002）^3
两边同除以N
左边：(2000x^2+2001y^2+2000z^2)/N=2000x^2/N+2001y^2/N+2000z^2/N
=2000x^2/2000x^3+2001y^2/2001y^3+2000z^2/2002z^3
=1/x+1/y+1/z
右边：（3次√2000+3次√2001+3次√2002）^3/N
=[(3次√2000/3次√N)+(3次√2001/3次√N)+(3次√2002/3次√N)]^3
=[(3次√2000/3次√2000x^3)+(3次√2001/3次√2001y^3)+(3次√2002/3次√2002z^3)]
=(1/x+1/y+1/z)^3
即是 1/x+1/y+1/z=(1/x+1/y+1/z)^3
化简 (1/x+1/y+1/z)^2=1
因为 2000x^3=2001y^3=2002z^3且xyz大于0
所以 x>0,y>0,z>0
所以 1/x+1/y+1/z=1
得证