dot product and norm "one"

Question

Let $\{x_i\}_{i=1}^{n}$ and $\{y_i\}_{i=1}^{n}$ two positive sequences, the first one is monotonic, the second one is strictly increasing .

I noticed that in many cases if $\{x_i\}_{i=1}^{n}$ is increasing $$\frac{\frac{1}{n}\sum_{i=1}^{n}{x_iy_i}}{\left(\frac{1}{n}\sum_{i=1}^{n}{x_i}\right)\left(\frac{1}{n}\sum_{i=1}^{n}{y_i}\right)}\ge1$$

and if $\{x_i\}_{i=1}^{n}$ is decreasing $$\frac{\frac{1}{n}\sum_{i=1}^{n}{x_iy_i}}{\left(\frac{1}{n}\sum_{i=1}^{n}{x_i}\right)\left(\frac{1}{n}\sum_{i=1}^{n}{y_i}\right)}\le1$$

I am very skeptical about the veracity of this assertion, but I cannot prove it, any help or directions would be welcome.

Answer 1

It is true indeed.

First let us claim the following Lemma ($\bullet$ will denote $\geq$ when $(x_i)$ is increasing and $\leq$ otherwise) :

For all $n\geq 2$ $$ \sum_{i=2}^nx_i\sum_{l=1}^{i-1}(y_i-y_l) \ \bullet \ \sum_{i=1}^{n-1}x_i\sum_{k=i+1}^{n}(y_k-y_i) $$

Assuming for the moment this Lemma, \begin{align*} & \sum_{i=2}^nx_i\sum_{l=1}^{i-1}(y_i-y_l) \ \bullet \ \sum_{i=1}^{n-1}x_i\sum_{k=i+1}^{n}(y_k-y_i) \\ \Leftrightarrow \quad & \sum_{i=2}^n (i-1)x_iy_i -\sum_{i=2}^nx_i\sum_{l=1}^{i-1}y_l \ \bullet \ \sum_{i=1}^{n-1}x_i \sum_{k=i+1}^{n}y_k-\sum_{i=1}^{n-1}(n-1-i)x_iy_i \\ \Leftrightarrow \quad & \sum_{i=2}^n (i-1)x_iy_i -\sum_{i>j}^nx_i y_j \ \bullet \ \sum_{i<j} x_i y_j-\sum_{i=1}^{n-1}(n-1-i)x_iy_i \\ \Leftrightarrow \quad & \sum_{i=2}^n (i-1)x_iy_i+ \sum_{i=1}^{n-1}(n-1-i)x_iy_i \ \bullet \ \sum_{i<j} x_i y_j+\sum_{i>j}^nx_i y_j \\ \Leftrightarrow \quad & (n-1)\sum_{i=1}^n x_iy_i \ \bullet \ \sum_{i\neq j}^nx_i y_j \\ \Leftrightarrow \quad & n\sum_{i=1}^n x_iy_i \ \bullet \ \sum_{i, j}^nx_i y_j \\ \Leftrightarrow \quad & n\sum_{i=1}^n x_iy_i \ \bullet \ \sum_{i=1}^nx_i \sum_{i=1}^ny_j \\ \Leftrightarrow \quad & \frac{\frac1n\sum_{i=1}^n x_iy_i}{\frac1n\Big(\sum_{i=1}^nx_i\Big)\frac1n\Big( \sum_{i=1}^ny_j\Big) } \ \bullet \ 1 \end{align*}

Proof of the Lemma: For $n=2$,

$x_2(y_2-y_1) \ \bullet \ x_1(y_2-y_1) \quad \Leftrightarrow \quad x_2 \ \bullet \ x_1$

Let us suppose the assumption true for $n$,

since $x_{n+1}\sum_{l=1}^n(y_{n+1}-y_l)\ \bullet \ \sum_{i=1}^nx_i(y_{n+1}-y_i)$, we have \begin{align*} \sum_{i=2}^{n+1}x_i\sum_{l=1}^{i-1}(y_i-y_l)&= \sum_{i=2}^nx_i\sum_{l=1}^{i-1}(y_i-y_l) + x_{n+1}\sum_{l=1}^n(y_{n+1}-y_l) \\ & \bullet \ \sum_{i=1}^{n-1}x_i\sum_{k=i+1}^{n}(y_k-y_i) +\sum_{i=1}^nx_i(y_{n+1}-y_i) \\ &= \sum_{i=1}^{n-1}x_i\sum_{k=i+1}^{n}(y_k-y_i) +\sum_{i=1}^{n-1}x_i(y_{n+1}-y_i)+x_n(y_{n+1}-y_n) \\ &=\sum_{i=1}^{n}x_i\sum_{k=i+1}^{n+1}(y_k-y_i) \end{align*} so $$\sum_{i=2}^{n+1}x_i\sum_{l=1}^{i-1}(y_i-y_l) \ \bullet \ \sum_{i=1}^{n}x_i\sum_{k=i+1}^{n+1}(y_k-y_i)$$ which concludes the proof.

Answer 2

You are looking for Chebyshev's sum inequality, quite useful though perhaps not as popular as AM-GM, Jensen and Cauchy Schwarz. The link has proofs, including a one liner if you are familiar with Rearrangement Inequality.