Positive definite quadratic form $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ Is $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ positive definite?

Approach:
The matrix of this quadratic form can be derived to be the following
$$M := \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} & \cdots & \frac{1}{2} \\
\frac{1}{2} & 1 & \frac{1}{2} & \cdots & \frac{1}{2} \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \cdots & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \cdots & 1 \\
\end{pmatrix}$$
It suffices to show that $\operatorname{det}M > 0$, then the claim follows.
Any hints how to show the positivity of this determinant?

The eigenvalues of an $n$ by $n$ matrix consisting entirely of 1's are $n$ and $0.$ An eigenvector for $n$ can be all entries 1. $0$ has multiplicity $n-1,$ for $1 \leq i \leq n-1$ take an eigenvector to have mostly 0's, but $-1$ at position $i$ and 1 at position $n.$  Adding the identity matrix makes the eigenvalues $n+1$ and $1.$ Dividing by 2 makes the eigenvalues $\frac{n+1}{2}$ and $\frac{1}{2}.$ 
It took me a few minutes to check, but the doubled version of this, with all 2's on the main diagonal and all 1's elsewhere, is isometric to the root lattice $A_n$ which comes from a Lie algebra. See A3, A4, A5, A6, A7 and so on.
