How to prove expression greater than 0 Prove that the following expression
$$6 \left( x_1^2 + \dots + x_n^2 \right) - 4 \left( x_1 x_2 + x_2 x_3 + \dots + x_{n-1} x_n \right) + 2 \left(x_1 x_n + x_2 x_{n-1} + \dots + x_{\frac{n+1}{2}−1} x_{\frac{n+1}{2}+1}\right)$$
is greater than $0$ for any vector $v = (x_1, \dots , x_n) \neq 0$ and for any odd $n$. 
Obviously, for $v = 0$ the expression is equal to $0$, and it also seems clear that the expression is indeed greater than $0$ for any $v \neq 0$, however I'm not able to put together a proof of this. 
Thank you for any help.
 A: One can rewrite the expression at hand in matrix form.
$$\begin{bmatrix}
x_1\\ 
x_2 \\ 
\vdots \\ 
\vdots \\
x_{n-1}\\
x_n\end{bmatrix}^T
\begin{bmatrix} 
6 & -2 & 0 & \cdots & 0 & 0 & 1\\
-2 & 6 & -2 & \cdots & 0 & 1 & 0\\
0 & -2 &  6 & \cdots & 1 & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
0 & 0 & 1 & \cdots & 6 & -2 & 0\\
0 & 1 & 0 & \cdots & -2 & 6 & -2\\
1 & 0 & 0 & \cdots & 0  & -2 & 6
\end{bmatrix}
\begin{bmatrix}
x_1\\ 
x_2 \\ 
\vdots \\ 
\vdots \\
x_{n-1}\\
x_n\end{bmatrix}
$$
One will notice the $n \times n$ matrix in the middle is real and symmetric.
This means all its eigenvalues are real.
This $n \times n$ matrix is composed of 4 pieces, a diagonal with all entries $6$. a super and a sub diagonals with entries $-2$ and an antidiagonal with entries $1$.
If one compute the row sums (i.e. the sum of absolute values of non-diagonal elements) on the rows, one will find they are at most $5$.
By Gershgorin circle theorem, the eigenvalues of this matrix will be contained in the disc $|z - 6| \le 5$ in complex plane. Since we know the eigenvalues are real, they are positive and $\ge 1$. This means that $n \times n$ matrix is positive definite and the expression is positive unless all $x_k = 0$.
A: Because for all odd $n$ and $x_{n+1}=x_1$ we obtain $$6\sum\limits_{k=1}^nx_k^2-4\sum_{k=1}^nx_kx_{k+1}+2\sum_{k=1}^{\frac{n-1}{2}}x_kx_{n-k+1}=$$
$$=2\sum_{k=1}^n(x_k-x_{k+1})^2+\sum_{k=1}^{\frac{n-1}{2}}(x_k-x_{n-k+1})^2+\sum_{k=1}^nx_k^2+x_{\frac{n+1}{2}}^2\geq0.$$
