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Say $(a_1, a_2, ..., a_n)$ and $(b_1, b_2, ..., b_n)$ are two sequences of increasing or decreasing positive real numbers.
Now, suppose we replace $(b_1, b_2, ..., b_n)$ by $(a_1^{-1}, a_2^{-1}, ... , a_n^{-1})$.
By Cauchy-Schwarz or AM-HM, $$ \left(\sum_{i=1}^n{a_i}\right)\left(\sum_{i=1}^n{a_i^{-1}}\right) \geq n^2 $$ In contrast, Chebyshev Inequality affirms that$-$ $$ \left(\sum_{i=1}^n{a_i}\right)\left(\sum_{i=1}^n{a_i^{-1}}\right) \leq n^2 $$ I found this result while playing around with Inequalities, and now I'm stuck.
How can this happen that one standard result states some other false?
I know that the first Inequality is definitely true, but what happens with the second one?

Answers are appreciated. Thanks!

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  • $\begingroup$ @Chrystomath How do you say that? I used it for positive real numbers. $\endgroup$ Aug 20, 2020 at 11:24
  • $\begingroup$ How did you get $n^2$ in the Chebyshev inequality? You have to keep both sequences in increasing order. $\endgroup$ Aug 20, 2020 at 11:28
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    $\begingroup$ Chebyshev says: $(\sum_i a_i)(\sum_i a_i^{-1})/n\le a_1/a_n+\cdots+a_n/a_1$. $\endgroup$ Aug 20, 2020 at 11:34
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    $\begingroup$ If $(a_1, a_2, ..., a_n)$ are in increasing order then $(a_1^{-1}, a_2^{-1}, ... , a_n^{-1})$ are in decreasing order. The inequality sign in Chebyshev's sum inequality reverses if the two sequences are in opposite order. So you get he same result as with Cauchy-Schwartz or with AM-HM. $\endgroup$
    – Martin R
    Aug 20, 2020 at 11:37
  • $\begingroup$ Okay, You solved my problem. Add the answer and I will upvote and accept. $\endgroup$ Aug 20, 2020 at 11:38

1 Answer 1

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Chebyshev's sum inequality states that $$ {1 \over n}\sum _{{k=1}}^{n}a_{k}\cdot b_{k}\geq \left({1 \over n}\sum _{{k=1}}^{n}a_{k}\right)\left({1 \over n}\sum _{{k=1}}^{n}b_{k}\right) $$ if $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ are both increasing (or both decreasing), and the reverse inequality $$ {1 \over n}\sum _{{k=1}}^{n}a_{k}\cdot b_{k}\leq \left({1 \over n}\sum _{{k=1}}^{n}a_{k}\right)\left({1 \over n}\sum _{{k=1}}^{n}b_{k}\right) $$ if the sequences are in opposite order (one increasing and the other decreasing).

With $(b_1, b_2, ..., b_n) =(a_1^{-1}, a_2^{-1}, ... , a_n^{-1})$ and increasing (or decreasing) $a_j$ we have the second case, which gives $$ 1 \le \left({1 \over n}\sum _{{k=1}}^{n}a_{k}\right)\left({1 \over n}\sum _{{k=1}}^{n}\frac{1}{a_{k}}\right) $$ i.e. the same result as Cauchy-Schwarz or the AM-HM inequality.

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