设2002x^3=2003y^3=2004z^3,xyz＞0,且“2002x^2+2003y^2+2004z^2”这整个

设2002x^3=2003y^3=2004z^3=k>0
则
2002x^2=k/x
2003y^2=k/y
2004z^2=k/z
2002=k/x^3
2003=k/y^3
2004=k/z^3
因为x,y,z>0 ,k>0
（√2002x^2+2003y^2+2004z^2)^(1/3)=2002^(1/3)+2003^(1/3)+2004^(1/3)
(k/x+k/y+k/z)^(1/3)=(k/x^3)^(1/3)+(k/y^3)^(1/3)+(k/z^3)^(1/3)
(1/x+1/y+1/z)^(1/3)=1/x+1/y+1/z
所以1/x+1/y+1/z+1