find the minimum of the positive real value $c$ such $x_{1}+x_{2}+\cdots+x_{n}=1$ Given the odd  positive integer $n>1$, find the minimum positive real value of $c$ such that for all $x_{i}\in \Bbb R$ with
$$x_{1}+x_{2}+\cdots+x_{n}=1,$$ it holds that
$$c\left(\sum_{i=1}^{n}x^2_{i}\right)^3\ge \left(\sum_{i=1}^{n}|x_{i+1}-x_{i}|\right)^2\left(\sum_{1\le i<j\le n}(x_{i}-x_{j})^2\right)$$
where $x_{n+1}=x_{1}.$
My attempt: since
$$\sum_{1\le i<j\le n}(x_{i}-x_{j})^2=n\sum_{i=1}^{n}x^2_{i}-(\sum_{i=1}^{n}x_{i})^2=n\sum_{i=1}^{n}x^2_{i}-n$$
and
$$(\sum_{i=1}^{n}|x_{i+1}-x_{i}|)^2=\sum_{i=1}^{n}(x_{i+1}-x_{i})^2+2\sum_{1\le i<j\le n}|x_{i+1}-x_{i}||x_{j+1}-x_{j}|=2\sum_{i=1}^{n}x^2_{i}-2\sum_{i=1}^{n}x_{i+1}x_{i}+2\sum_{1\le i<j\le n}|x_{i+1}-x_{i}||x_{j+1}-x_{j}|$$
Then I can't ,maybe this problem is from a Integral discrete it?
 A: We can show that River Li’s bound $c_n\ge \frac{16}{27}(n-1)n^2=b_n$ is tight as follows. 
Given any real $x_1,\dots x_n$ with $\sum x_i=1$ for each $i$ put $y_i=x_i-\tfrac 1n$. Then $x_i-x_j=y_i-y_j$ for each $i$ and $j$, $\sum y_i=0$, and  
$$\sum x_i^2=\sum \left(y_i+\frac 1n\right)^2=\sum y_i^2+\frac 2n y_i+\frac 1{n^2}=$$ $$
\sum y_i^2+ \frac 2n\left(\sum y_i\right)+n\frac 1{n^2}=\sum y_i^2+\frac 1{n}.$$
Then 
$$\sum_{1\le i<j\le n}(x_{i}-x_{j})^2=n\sum_{i=1}^{n}x^2_{i}-1=n\sum y_i^2=nY\ge 0.$$
Thus we have to show that 
$$b_n(Y+\tfrac 1n)^3\ge n\left(\sum_{i=1}^{n}|y_{i+1}-y_{i}|\right)^2Y.$$
For this we need the following auxiliary result. 
For each natural $k$ and each point $z=(z_1,\dots, z_k)\in \Bbb R^k$ let $\|z\|=\sqrt{\sum_{i=1}^k z_i^2}$, $B_k=\{z\in\Bbb R^k: \|z\|\le 1\}$ be the unit ball in the space $\Bbb R^k$, and $f:B_k\to\Bbb R$ be a function such that $$f(z)=\sum_{i=1}^{k}|z_{i+1}-z_{i}|$$
for each $z\in B_k$, where $z_{k+1}=z_1$. Since the set $B_k$ is compact and the function $f_k$ is continuous, it attains its maximum $M_k$ at some point $t=(t_1,\dots,t_k)\in B_k$. Clearly, $\sum_{i=1}^{n}|y_{i+1}-y_{i}|\le \sqrt{Y}M_k$. 
By the inequality between arithmetic and quadratic means, we have 
$$M_k=\sum_{i=1}^{k}|t_{i+1}-t_{i}|\le \sum_{i=1}^{k}|t_{i+1}|+|t_{i}|=$$ $$2\sum_{i=1}^{k}|t_i|\le 
2\sqrt{k}\sqrt{\sum_{i=1}^{k}t_i^2}=2\sqrt{k}\|t\|\le 2\sqrt{k}.$$ 
It is easy to check that if $k$ is even then the equality above is attained when $t_i=\tfrac {(-1)^i}{\sqrt{k}}$ for each $i$.  
Now assume that $k>1$ is odd. Suppose to the contrary that all $t_i$ are non-zero. We claim that there exists a natural $i$ such that both $t_i$ and $t_{i+1}$ (in the circular order) have the same sign. Indeed, otherwise the sign of each $t_i$ is $(-1)^{i-1}\operatorname{sign} t_1$. Since $k$ is odd, $t_1$ and $t_k$ have the same sign, a contradiction. Let $t_j,\dots, t_l$ be a longest circular subsequence of consecutive $t_i$’s with the same sign. Let $t_m$ be a number in this subsequence with  the smallest absolute value. Let a point $t’\in\Bbb R^k$ has the same coordinates as $t$, but only its $m$-th coordinate is zero, and $$\lambda=\frac {\|t\|}{\|t’\|}>1.$$ Then $\|\lambda t’\|=\|t\|\le 1$, so $\lambda t’\in B_k$. On the other hand, 
it is easy to check that $f(t’)\ge f(t)$, so $f(\lambda t’)=\lambda f(t’)>f(t)$, a contradiction with the maximality of $f(t)$. Thus there exists $m$ such that $t_m=0$. Then similarly to the above case of even $k$ we can show that 
$$f(t)\le  2\sum_{i=1}^{k}|t_i|\le 2\sqrt{k-1}\sqrt{\sum_{i=1}^{k}t_i^2}=2\sqrt{k-1}\|t\|\le 2\sqrt{k-1}.$$
It is easy to check that if the equality above is attained when $t_i=\tfrac {(-1)^i}{\sqrt{k-1}}$ for each $i<k$ and $t_k=0$. Thus $M_k=2\sqrt{k-1}$.  
Since $n$ is odd, $$\left(\sum_{i=1}^{n}|y_{i+1}-y_{i}|\right)^2\le YM_n^2=4Y(n-1).$$
So it remains to show that 
$b_n(Y+\tfrac 1n)^3\ge 4(n-1)nY^2$
$\frac{16}{27}(n-1)n^2(Y+\frac 1n)^3\ge 4(n-1)nY^2$
$4n(Y+\frac 1n)^3\ge 27Y^2$
$4(nY+1)^3\ge 27n^2Y^2$
Put $y=nY\ge 0$. We have to show that 
$4(y+1)^3\ge 27y^2$
$4y^3-15y^2+12y+4\ge 0$
$(y-2)^2(4y+1)\ge 0$.
A: With numerical searching, I guess that the best constant $c_n$ is $\frac{16}{27}(n-1)n^2$. 
Some values: $c_3 = 32/3, \ c_5 = 1600/27, \ c_7 = 1568/9, \ c_9 = 384, \ c_{11} = 19360/27, \cdots$
Are there any counterexamples (by computer)?
Details: We have
$$c_n = \sup_{x_1 + x_2 + \cdots + x_n = 1}
\frac{\left(\sum_{i=1}^{n}|x_{i+1}-x_{i}|\right)^2\left(n\sum_{i=1}^{n}x^2_{i} -1\right)}{\left(\sum_{i=1}^{n}x^2_{i}\right)^3}.$$
Let $(x_1, x_2, \cdots, x_n) = (a, b, a, b, \cdots, a, b, \frac{1}{n})$ where
$$a = \frac{1}{n} + \frac{1}{\sqrt{\frac{n(n-1)}{2}}}, \quad b = \frac{1}{n} - \frac{1}{\sqrt{\frac{n(n-1)}{2}}}.$$
We have
$$\frac{\left(\sum_{i=1}^{n}|x_{i+1}-x_{i}|\right)^2\left(n\sum_{i=1}^{n}x^2_{i} -1\right)}{\left(\sum_{i=1}^{n}x^2_{i}\right)^3} = \frac{16}{27}(n-1)n^2.$$
A: As seen in Alex Ravsky's answer, the desired minimum is the same as the maximum of the expression
$$
(nY+1)^{-3}n^4Yf^2(y)
$$
subject to $y_1+\cdots+y_n = 0$. For this, let us define the function $g(x) := \frac{x^2}{(x+1)^3}$. It has a global maximum at $x=2$ with $g(2) = \frac 4{27}$. As shown by Alex, we have
$$
(nY+1)^{-3}n^4Yf^2(y)\le 4(n-1)(nY+1)^{-3}n^4Y^2 = 4(n-1)n^2g(nY)\le\frac{16}{27}(n-1)n^2.
$$
On the other hand, set $\tau =\sqrt{2/n}$ and choose $y\in\mathbb R^n$ such that $y_k = \frac {(-1)^k\tau}{\sqrt{n-1}}$ for $k=1,\ldots,n-1$ and $y_n = 0$. Then $Y = \tau^2 = 2/n$ and $y_1+\ldots+y_n = 0$. Plugging $y$ into our expression gives
$$
\frac{2n^3}{27}f(y)^2 = \frac{2n^3}{27}\cdot 4(n-1)Y = \frac{16}{27}(n-1)n^2.
$$
