已知x>0, y>0, z>0, 且3x+2y+z=1, 求

1/(3x)+1/(2y)+1/(z)= [1/(3x) + 1/(2y) + 1/(z)] * 1= [1/(3x) + 1/(2y) + 1/(z)] * (3x + 2y + z)= 3x/(3x) + 3x/(2y) + 3x/z + 2y/(3x) + 2y(2y) + 2y/z + z/(3x) + z/(2y) + z/z= 1 + 3x/(2y) + 3x/z + 2y/(3x) + 1 + 2y/z + z/(3x) + z/(2y) + 1= 3 + [3x/(2y) + 2y/(3x)] + [3x/z + z/(3x)] + [2y/z + z/(2y)]>= 3 + 2√[3x/(2y) * 2y/(3x)] + 2√[3x/z * z/(3x)] + 2√[2y/z * z/(2y)]= 3 + 2 + 2 + 2= 91/(3x)+1/(2y)+1/(z)之最小值是_9___.


2010-09-21 19:59:40 補充：
打得好辛苦,請不要移除問題,謝謝!

1/(3x) + 1/(2y) + 1/(z)
= (3x+2y+z)/(3x) + (3x+2y+z)/(2y) + (3x+2y+z)/(z)
=...