# Can you prove the rearrangement inequality using Cauchy-Schwarz?

For $$n$$ a positive integer, let $$(a_1 a_2 ,\ldots, a_n)$$ and $$(b_l, b_2 ,\ldots, b_n)$$ be two (not necessarily distinct) permutations of $$(1,2, ... ,n)$$. Find sharp lower and upper bounds for $$a_1b_1 + \ldots + a_nb_n$$

My upper bound and lower bounds are (resp):

$$\sqrt{\sum{a_i^2b_i}\sum {b_i}}$$

$$\dfrac{\bigg(\sum \sqrt{a_ib_i}\bigg)^2}{\sum \sqrt{a_i}}$$

I would love to know if we can improve on these bounds. Moreover, I was hoping that I can prove the rearrangement inequality from this but I don't think that's possible since Cauchy-Schwarz doesn't care about the order of the inner product terms.

Using Cauchy-Schwarz we get:

\begin{align} \sum_{i=1}^{n}{a_{i}b_{i}}&\leq\sqrt{\left(\sum_{i=1}^{n}{a_{i}^{2}}\right)\left(\sum_{i=1}^{n}{b_{i}^{2}}\right)}\\ &\leq\sum_{i=1}^{n}{i^{2}} \end{align}

Equality occurs when $$a_{i}=b_{i}$$ i.e. when $$a_{i}$$ and $$b_{i}$$ are in the same order. This align with upper bound and equality condition of rearrangement inequality.

$$\rule{8cm}{0.4pt}$$

We now need to find the lower bound. Define $$c_{i}$$ such that $$b_{i}=n+1-c_{i}$$. Easy to see that $$\left(c_{1},...,c_{n}\right)$$ is also a permutation of $$\left(1,...,n\right)$$.

\begin{align} \sum_{i=1}^{n}{a_{i}b_{i}}&=\left(n+1\right)\sum_{i=1}^{n}{a_{i}}-\sum_{i=1}^{n}{a_{i}c_{i}}\\ &\geq\left(n+1\right)\sum_{i=1}^{n}{i}-\sum_{i=1}^{n}{i^{2}} \end{align}

Equality occurs when $$a_{i}=c_{i}$$ i.e. when $$a_{i}$$ and $$b_{i}$$ are in reverse order. Once again, align with rearrangement inequality.

• I guess in this special case it is possible to obtain a bound that aligns with the RI
– user675768
Commented Aug 29, 2020 at 2:44