Talk:Chebyshev's sum inequality

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)

"Similarly, if a1 <= a2 ..., and b1 >= b2 ..." shouldn't this be "b1 <= b2 ..."? 77.127.118.186 (talk) 11:53, 31 August 2018 (UTC)