pure maths .........

let ai = 1+ti as the hint told. where Sum[ai]= Sum[1+ti] = n +Sum[ti] =n => Sum[ti]=0

Sum[ ai^2] = Sum [ (1+ti)^2] = Sum [ 1+2 ti +ti^2] = n + 2 Sum[ti] +Sum[ti^2] = n +Sum[ti^2]

since ti^2 are all positive, so, Sum[ti^2] >0

Thus Sum[ ai^2] > n

the equality hold iff Sum[ti^2]=0 iff ti =0 iff ai=1

2008-01-02 13:59:57 補充：
Hey, i saw the proof from down stair, the first statement is wrong, coz ai can be smaller than 1, then, ai^2 < aiand this is the difficulty of the question. you cannot assume all ai >= 1, else, Sum[ai] >=n

應該係-1

please see the image below
http://i172.photobucket.com/albums/w5/fama411/241173114.jpg

我冇用到hint
有錯的話請指責