Talk:Chebyshev's sum inequality

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You have new messages (last change).

I changed the statement of the inequality from

n \sum_{k=1}^n a_kb_k \geq \left(\sum_{k=1}^n a_k\right)\left(\sum_{k=1}^n b_k\right)

to

{1\over n} \sum_{k=1}^n a_kb_k \geq \left({1\over n}\sum_{k=1}^n a_k\right)\left({1\over n}\sum_{k=1}^n b_k\right),

to conform to the final line in the proof, and also to correspond more closely to the "continuous version" at the end of the article. Gabn1 09:19, 27 September 2006 (UTC)

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