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?

We can more generally show that the $n\times n$ determinant of $$\left|\begin{array}{cccc}a & b & \cdots &  b \\ c & a &  \ddots & \vdots\\ \vdots & \ddots & \ddots  & b \\ c & \cdots & c & a\end{array}\right|$$
is $\left\{\begin{array}{ll}\frac{1}{c-b}(c(a-b)^n-b(a-c)^n) & \text{ if }b\neq c \\ (a-b+nb)(a-b)^{n-1}& \text{ if } b=c\end{array}\right.$
So in your case, the determinant is $\frac{n+1}{2^n}>0$.
Proof :
Let $$H(x)=\left|\begin{array}{cccc}a-x & b-x & \cdots &  b-x \\ c-x & a-x &  \ddots & \vdots\\ \vdots & \ddots & \ddots  & b-x \\ c-x & \cdots & c-x & a-x\end{array}\right|$$
Do $L_2\leftarrow L_2-L_1,\ldots,L_n\leftarrow L_n-L_1$ :
$$H(x)=\left|\begin{array}{ccc}a-x & b-x & \cdots & \cdots & \cdots & b-x \\ \\ c-a & a-b & 0&\cdots&\cdots& 0\\\vdots & c-b & \ddots&\ddots&&\vdots\\\vdots&\vdots & \ddots&\ddots&\ddots&\vdots\\\vdots&\vdots&&\ddots&\ddots&0\\c-a&c-b&\cdots&\cdots&c-b&a-b\end{array}\right|$$
and $C_2\leftarrow C_2-C_1,\ldots,C_n\leftarrow C_n-C_1$ :
$$H(x)=\left|\begin{array}{cccccc}a-x&b-a&\cdots&\cdots&\cdots&b-a\\c-a&2a-b-c&a-c&\cdots&\cdots&a-c\\\vdots&a-b&\ddots&\ddots&&\vdots\\
\vdots&\vdots&\ddots&\ddots&\ddots&\vdots\\\vdots&\vdots&&\ddots&\ddots&a-c\\c-a&a-b&\cdots&\cdots&a-b&2a-b-c\end{array}\right|$$
So it exists $\alpha$ and $\beta$ such as $H(x)=\alpha x+\beta$.
If $b\neq c$ we can evaluate $H(b)$ and $H(c)$ with the first writing of $H(x)$ : $$\left\{\begin{array}{c}\alpha b+\beta=(a-b)^n \\ \alpha c+\beta=(a-c)^n\end{array}\right.$$
So you get $\alpha$ and $\beta$, and your determinant is $H(0)=\frac{1}{c-b}(c(a-b)^n-b(a-c)^n)$. 
If $b=c$ : your determinant is $\lim_{c\rightarrow b}\frac{1}{c-b}(c(a-b)^n-b(a-c)^n)=-f'(b)$ where $f(x)=(c+b-x)(a-x)^n$.
We can easily improve this proof to the case where the elements on the diagonal are not equal
