<s>[INST] 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?
 [/INST] One possible proof is to use the fact that for triangular $n\times n$ matrix $A$ it holds
$$\det(A) = \prod_{i=1}^n a_{i,i}$$
An example
$$\begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & 1 & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & 1 \\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} \\
0 & \frac{3}{4} & \frac{1}{4} \\
0 & \frac{1}{4} & \frac{3}{4} \\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} \\
0 & \frac{3}{4} & \frac{1}{4} \\
0 & 0 & \frac{2}{3} \\
\end{pmatrix}
$$
Determinant of the last form is clearly positive. Try to generalize this result.
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