dot product and norm "one" 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.
 A: 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.
A: 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.
