find this inequality min (a) let $x_{k}\in R$,and $\displaystyle\sum_{k=1}^{n}x_{k}=1,n=2m+1,m\in N^{*}$,and
$a\left(\displaystyle\sum_{k=1}^{n}x^2_{k}\right)^3\ge\left(\displaystyle\sum_{k=1}^{n}|x_{k+1}-x_{k}|\right)^2\left(n\displaystyle\sum_{k=1}^{n}x^2_{k}-\left(\displaystyle\sum_{k=1}^{n}x_{k}\right)^2\right),x_{n+1}=x_{1}$
find the $\min{a}=?$
 A: Just a suggestion. :/
$a = \max_{x_1,\cdots,x_n}\dfrac{\left(\sum_{k=1}^{n-1}|x_{k+1}-x_k|\right)^2\left(n\sum_{k=1}^n x_k^2 - \left(\sum_{k=1}^n x_k\right)^2\right)}{\left(\sum_{k=1}^n x_k^2\right)^3}$
If we fix the values of $x_1,\cdots, x_n$, we can see we need to maximize the numerator by permute the $x_1,\cdots,x_n$, and the denom will not change.
Rearrange 
suppose initially $x_1\le x_2 \le \cdots, x_n$, we take a permutation $\pi : \vec{x}\rightarrow \vec{y}$, such that 
$y_1\le y_3\le y_5\cdots\le y_{2m+1}\le y_{2m}\le y_{2m-2}\le \cdots \le y_2$,
we shall say this (or reversed version) can maximize the numerator.
Actually, if you take any permutation to this, we can see this will decrease the sum.

Without loss of generality, we assume our $x_1,\cdots,x_n$ has been arranged in this way.
Which means
$x_1\le x_3\le x_5\cdots\le x_{2m+1}\le x_{2m}\le x_{2m-2}\le \cdots \le x_2$,
Then say the gaps between them is $d_1,\cdots d_{n-1}$, where $d_i$ are non-negative.
$x_3 = x_1+d_1$,
$x_5 = x_1 + d_1 +d_2$,
$\dots$,
$x_4 = x_1 + d_1 + d_2 + \dots +d_{2m-1}$
$x_2 = x_1 + d_1 +d_2 +\dots + d_{2m-1}+d_{2m}$ ,
And $$(2m+1)x_1 + \sum_{i=1}^{2m} (2m+1-i)d_i = 1$$
thus $$x_1 = \dfrac{1-\sum_{i=1}^{2m} (2m+1-i)d_i}{2m+1}\le \dfrac{1}{2m+1}$$ and 
$$\sum_{i=1}^{2m} (2m+1-i)d_i = 1-(2m+1)x_1$$
$$\sum_{k=1}^{n-1} |x_{k+1}-x_k| = d_1 + 2d_{2m} + 3d_2+4d_{2m-1}+\cdots+ (2m)d_{m+1}$$
$$\sum_{k=1}^n x_k^2 = x_1^2+(x_1+d_1)^2+\cdots+(x_1+d_1+\cdots+d_{2m})^2 = (2m+1)x_1^2 + 2x_1\sum_{i=1}^{2m}(2m+1-i)d_i + \sum_{l=1}^{2m}(\sum_{k=1}^l d_k)^2 $$
Then all the arguments are with $d_1,d_2,\cdots,d_{2m}$, the only requirements for $d_i$ is non-negative.
