inequality

By the Cauchy Schwarz ineq, we have:
[Σ(i = 1 → n) ai4][Σ(i = 1 → n) bi4] = [Σ(i = 1 → n) (ai2)2][Σ(i = 1 → n) (bi2)2]
>= [Σ(i = 1 → n) ai2bi2]2
and [Σ(i = 1 → n) ci4][Σ(i = 1 → n) di4] >= [Σ(i = 1 → n) ci2di2]2
Hence, applying CS-ineq, we have:
[Σ(i = 1 → n) ai4][Σ(i = 1 → n) bi4][Σ(i = 1 → n) ci4][Σ(i = 1 → n) di4] >= [Σ(i = 1 → n) ai2bi2]2[Σ(i = 1 → n) ci2di2]2
= {[Σ(i = 1 → n) (aibi)2][Σ(i = 1 → n) (cidi)2]}2
>= {[Σ(i = 1 → n) (aibicidi)]2}2
= [Σ(i = 1 → n) (aibicidi)]4
Then by setting ai4 = ui3, bi4 = vi3, ci4 = ti3 and ai4 = uiviti, we have:
[Σ(i = 1 → n) ui3][Σ(i = 1 → n) vi3][Σ(i = 1 → n) ti3][Σ(i = 1 → n) uiviti] >= [Σ(i = 1 → n) uiviti]4
[Σ(i = 1 → n) ui3][Σ(i = 1 → n) vi3][Σ(i = 1 → n) ti3 >= [Σ(i = 1 → n) uiviti]3
Therefore, by replacing the dummy variables ui3, vi3, and ti3 by ai3, bi3, and ci3 respectively, we have:
[Σ(i = 1 → n) ai3][Σ(i = 1 → n) bi3][Σ(i = 1 → n) ci3 >= [Σ(i = 1 → n) aibici]3

2009-07-18 20:35:48 補充：
In fact for the power 3 one, it is only true for the positive nos.
Taking:
a1 = -1, a2 = -4
b1 = -2, b2 = -5
c1 = -3, c2 = -6
The result is:
L.H.S. = -2100735
R.H.S. = -2100376

答案正是http://hk.knowledge.yahoo.com/question/article?qid=6908032501912所示的內容。

想不到飛天魏國大將軍張遼的興趣題竟然變成了Pure Maths題！

In the 2 solutions provided, x/v must be positive. But the question said a,b,c and d are real: so can be negative. So you guys have proved the case for a,b,c and d are positive real numbers?

2009-07-18 20:44:37 補充：
So indeed the question should be revised to mention positive real numbers only.