How find the minmum of the value $f=x^2_{1}+x^2_{2}+\cdots+x^2_{n}-x_{1}x_{2}-x_{2}x_{3}-\cdots -x_{n}x_{1}$ 
Assume the $n$ is give postive integer,let the pairwise distinct integers $x_{1},x_{2},\cdots,x_{n},n\ge 0$. find the  minmum of the value
  $$f=x^2_{1}+x^2_{2}+\cdots+x^2_{n}-x_{1}x_{2}-x_{2}x_{3}-\cdots -x_{n}x_{1}$$

I know 
$$2f=\sum_{i=1}^{n}(x_{i}-x_{i+1})^2 =\sum_{i=1}^{n-1}(x_{i+1}-x_{i})^2+(x_{n}-x_{1})^2\ge (n-1)+(n-1)^2=n^2-n?$$
I fell this reslut is not true,so How to slove it?
 A: With $y_i=x_{i+1}-x_i$ (and $y_n=x_1-x_n$), you showed that
$$\tag12f=\sum_{i=1}^n y_i^2.$$
If there is an integer $m$ with $\min x_i<m<\max x_i$, then replacing all $x_i>m$ with $x_i-1$ still leaves us with distinct integers, none of the summands in $(1)$ increases, at least two summands decrease. We conclude that in the minimal case, the $x_i$ are consecutive integers (in some order). Also, $(1)$ shows that replacing all $x_i$ with $x_i+d$ leaves $f$ unchanged. Thus we may assume that $x_1,x_2,\ldots, x_n$ is a permutation of $1,2,\ldots, n$.
Assume $(x_1,\ldots,x_n)$ is a minimizer. Pick $i,j$ with $j-i\ge 3$. Then the value of $(x_1,x_2,\ldots, x_i,x_{j-1},x_{j-2}\ldots, x_{i+1},x_j,x_{j+1},\ldots, x_n)$ (i.e., the terms between $x_i$ and $x_j$ reversed) differs from that of $(x_1,\ldots, x_n)$ by
$$ x_{j-1}x_i+x_{i+1}x_j-x_ix_{i+1}-x_{j-1}x_j=(x_{j-1}-x_{i+1})(x_i-x_j).$$
As this must be non-negative, we conclude that

$(2)$ In a minimizer, the sign of $x_{j-1}-x_{i+1}$ is the same as the sign of $x_j-x_i$ for $j\ge i+3$.

Assume that $\{x_1,\ldots x_r\}=\{1,\ldots r\}$. Assume $x_k=r+1$ with  $r+1< k<n$. 
Then $x_{r+1}>x_k$, hence by $(2)$, we have $r\ge x_r>x_{k+1}>r$, contradiction.
It follows that $x_{r+1}=r+1$ or $x_n=r+1$.
If additionally $x_1>x_r$ and $r<n-1$, it follows from $(2)$ that $x_{r+1}<x_n$ so that only the case $x_{r+1}=r+1$ remains.
As the sequence can be cyclically rearranged, we conclude by induction that it can be constructed as follows: Start with $1,2$, and then for $i=3,\ldots, n$, append $i$ to the end with the smaller number (which is $i-2$). Clearly, this leads to all odd numbers appended to the left of $1$ and all even numbers appended to the right of $2$.
We end up with a minimizer with $|x_{i+1}-x_i|=2$ (cyclically) for all $i$ except that $|x_{i+1}-x_i|=1$ for two $i$ (once near $1$ and once near $n$). By $(1)$, this leads to $2f=(n-2)\cdot 2^2+2\cdot 1^2$, or
$$ f=2n-3.$$
A: For the integers from $1$ to $n$ you can make two differences $1$ and all other differences $2$ by arranging the numbers as:- 
$$..., 5,3,1,2,4,6....$$ with $n$ eventually reached on one side or another.
This gives a total of $4n-6$. 
Note that there is no loss of generality in assuming that integers are from $1$ to $n$ because: 
they can be translated until the smallest $x_i$ is $1$; 
if the largest $x_i$ were greater than $n$ then we could simply reduce $2f$ by subtracting 1 from some of the larger $x_i$ without making two of the $x_i$ equal.
Proof
In general, we have two chains of numbers from $1$ to $n$ and therefore we have $n$ integers (the differences) 
such that$$d_1+d_2+...+d_n=2(n-1) \;\;\;\;(1)$$ and we wish to prove that  $$d_1^2+d_2^2+...+d_n^2\ge 4n-6.$$ 
We can assume the $d_i$ are non-negative. Suppose two of the $d_i$ were $a$ and $b$ where $b\ge a+2$. Then replacing $a$ and $b$ by $a+1$ and $b-1$ decreases the sum of squares. 
Therefore, for the minimum, the $d_i$ can only take two values which differ by $1$. From equation (1) these values must be $1$ and $2$ giving the above solution.
Therefore the minimum value of $f$ is $2n-3.$
