AL inequality

2010-01-07 6:14 am
here is the question:

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更新1:

神人.... 你係點諗到代咩係 lambda 同 x ? 做得多?

回答 (1)

2010-01-07 6:37 am
✔ 最佳答案

Let λr = br^3 / (Σbk^3)
λ1 + λ2 + λ3 + ... λn = (b1^3 + b2^3 + ... bn^3) / (Σbk^3) = 1 satisfying given condition
Let xr = ar/br
Sub into the given inequality,
[Σλx]^3 <= Σλx^3 (subscript omitted for clarity)
[Σ(br^3/Σ bk^3)(ar/br)]^3 <= Σ(br^3/Σ bk^3)(ar/br)^3
[Σ(br^3)(ar/br)]^3 / [Σ (bk)^3]^3 <= [Σ(br^3)(ar/br)^3] / Σ (bk)^3
[Σ(arbr^2)]^3 <= [Σ(ar)^3][Σ(bk)^3]^2
[Σ(arbr^2)]^3 <= [Σ(ar)^3][Σ(br)^3]^2

2010-01-07 00:29:12 補充:
都是經驗


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