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?

This can be solved by probability theory: Let your matrix be $\Sigma$ with diagonal elements all 1 and the off-diagonal elements all equal to $\rho$.
This is the variance-covariance matrix of a random vector $X$ where
all components have variance 1 and all pairs of distinct elements have correlation $\rho$, where $|\rho| \le 1$ is necessary for this having any hope of being a variance-covariance matrix (which implies the determinant is nonnegative)
 Are there any other restrictions? Yes, we can calculate the variance of the sum of the elements of $X$, which clearly must be positive:
$$\mbox{var}\left(\sum_{i=1}^n X_i\right) = 1^T\Sigma 1 = \sum_i \sum_j \Sigma_{ij} =n + n(n-1) \rho >0$$
(where 1 stands for the vector with all ones)
First cancel one factor $n$, then solve the inequality, you get
$\rho > -\frac{1}{n-1}$, and your value 1/2 clearly satisfies this!
