# Does Cauchy-Schwarz Inequality contradict Chebyshev's Inequality?

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!

• @Chrystomath How do you say that? I used it for positive real numbers. Aug 20 '20 at 11:24
• How did you get $n^2$ in the Chebyshev inequality? You have to keep both sequences in increasing order. Aug 20 '20 at 11:28
• Chebyshev says: $(\sum_i a_i)(\sum_i a_i^{-1})/n\le a_1/a_n+\cdots+a_n/a_1$. Aug 20 '20 at 11:34
• 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. Aug 20 '20 at 11:37
• Okay, You solved my problem. Add the answer and I will upvote and accept. Aug 20 '20 at 11:38

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.