Establish the inequality $\frac{1}{n}\sum_{i=1}^{n}x_{i}x_{n-i+1} < \left(\frac{1}{n}\sum_{i=1}^{n}x_{i}\right)^2$ Given a finite increasing sequence of real numbers $\{x_i\}_{i=1}^n$ consisting of at least two elements, how can we show that 
$$
\frac{1}{n}\sum_{i=1}^{n}x_{i}x_{n-i+1} < \left(\frac{1}{n}\sum_{i=1}^{n}x_{i}\right)^2
$$
without expanding. I attempted applying Holder's inequality but failed. Maybe I missed something.
 A: Hints:
1) the sum in LHS is by Rearrangement Inequality the lowest possible of form $x_ix_{\pi(i)}$ where $\pi$ is a permutation.
2) unless the $x_i$ are strictly increasing somewhere, equality is possible. 

P.S. For some more detail, note that $n\sum x_ix_{n−i+1}<\left(\sum x_i\right)^2$ follows from $n$ rearrangements summed together.  Each such rearrangement has an RHS of form $\sum x_i x_{i+j}$ where $j=0,1,...n−1$, and the sum in LHS is lower than each of them.  
A: This should be wrong.
For example for $x_i = 1$ this is an equality isn't it?
$$ \frac{1}{n} \sum \limits_{i=1}^{n} 1^2 = 1 =  1^2 = (\frac{1}{n} \ n)^2 = (\frac{1}{n} \sum \limits_{i=1}^{n} 1)^2 $$
A: Let
$$
\bar{x}=\frac1n\sum_{i=1}^nx_i\tag1
$$
If $x_i\le x_{i+1}$ and $x_1\lt x_n$, we can find a $k$ so that $x_k\le\bar{x}\lt x_{k+1}$. Then
$$
\bar{x}-x_{n-i+1}\ge0\iff i\ge n-k+1\tag2
$$
Therefore,
$$
\begin{align}
\left(\frac1n\sum_{i=1}^nx_i\right)^2-\frac1n\sum_{i=1}^nx_ix_{n-i+1}
&=\frac1n\sum_{i=1}^nx_i\bar{x}-\frac1n\sum_{i=1}^nx_ix_{n-i+1}\tag3\\
&=\frac1n\sum_{i=1}^nx_i\left(\bar{x}-x_{n-i+1}\right)\tag4\\
&=\frac1n\sum_{i=1}^n\left(x_i-x_{n-k+1}\right)\left(\bar{x}-x_{n-i+1}\right)\tag5
\end{align}
$$
Explanation:
$(3)$: apply $(1)$
$(4)$: distributive property
$(5)$: subtract $\frac1nx_{n-k-1}\sum\limits_{i=1}^n\left(\bar{x}-x_{n-i+1}\right)=0$
Applying $(2)$,
$$
\begin{align}
i\ge n-k+1&\implies\bar{x}-x_{n-i+1}\ge0\quad\text{and}\quad x_i-x_{n-k+1}\ge0\tag6\\
i\lt n-k+1&\implies\bar{x}-x_{n-i+1}\lt0\quad\text{and}\quad x_i-x_{n-k+1}\le0\tag7
\end{align}
$$
Thus, the terms in $(5)$ are non-negative. Furthermore, either the $i=1$ term or the $i=n$ term of $(5)$ is positive; that is, $\bar{x}$ cannot equal either $x_1$ or $x_n$, and $x_{n-k+1}$ cannot equal both $x_1$ and $x_n$.
Summarizing, if $x_i\le x_{i+1}$ and $x_1\lt x_n$,
$$
\frac1n\sum_{i=1}^nx_ix_{n-i+1}\lt\left(\frac1n\sum_{i=1}^nx_i\right)^2\tag8
$$
A: $$\frac{1}{n}\sum_{i=1}^{n}x_{i}x_{n-i+1} < \left(\frac{1}{n}\sum_{i=1}^{n}x_{i}\right)^2\iff\sum_{i=1}^{n}x_{i}x_{n-i+1} < \frac{1}{n}\left(\sum_{i=1}^{n}x_{i}\right)^2$$
When $n$ is odd one has $LHS=2(x_1x_n+\cdots+x_{\frac{n-1}{2}}x_{\frac{n+3}{2}})+x^2_{\frac{n+1}{2}}$
When $n$ is even one has $LHS=2(x_1x_n+\cdots+x_{\frac{n}{2}}x_{\frac{n+2}{2}})$
In both cases we have trivially that a sum of squares distinct of zero is greater than $0$.
$$0\lt n((x_n-x_1)^2+\cdots +(x_{\frac{n+3}{2}}-x_{\frac{n-1}{2}})^2)\text{ when n is odd               }\\0\lt n((x_n-x_1)^2+\cdots +(x_{\frac{n+2}{2}}-x_{\frac{n}{2}})^2)\text{ when n is even              }$$
