I have a matrix, given by
$$ \frac{1}{(1+x_1^2+x_2^2+\cdots+x_n^2)^{3/2}}\begin{bmatrix} x_{1}^2 & x_{1}x_2 & x_{1}x_3 & \dots & x_{1}x_n \\ x_{2}x_1 & x_{2}^2 & x_{2}x_3 & \dots & x_{2}x_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{n}x_1 & x_{n}x_2 & x_{n}x_3 & \dots & x_{n}^2 \end{bmatrix}, $$ where $x_i \in \mathbb{R}$. I want to show that this is positive (semi)definite. There are many methods, naturally. I could show the determinant of the all the minors are positive, but how? I could try to work out the eigenvalues, but how? I could try and show this matrix is diagonally dominant with positive diagonals, but how? I could simply apply the definition of positive definite-ness, and see if the resulting number is non-negative, but it is a mess and it's not immediate to me why the result is non-negative.
I appreciate any help, thank you.
Observe the following:
$$ A := \frac{1}{(1+x_1^2+x_2^2+\cdots+x_n^2)^{3/2}}\begin{bmatrix} x_{1}^2 & x_{1}x_2 & x_{1}x_3 & \dots & x_{1}x_n \\ x_{2}x_1 & x_{2}^2 & x_{2}x_3 & \dots & x_{2}x_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{n}x_1 & x_{n}x_2 & x_{n}x_3 & \dots & x_{n}^2 \end{bmatrix}, $$
can be equivalently written as
$$ A(x) := \frac{1}{g(x)} xx^T, \quad g(x) := (1+x_1^2+x_2^2+\cdots+x_n^2)^{3/2} > 0, \forall x \in \mathbb{R}^n $$ where $x = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n\end{pmatrix}$.
By the definition of positive (semi)definiteness, it suffices to prove that
$$ \begin{align*} z^T A(x) z &\geq 0, \forall z \in \mathbb{R}^n \Leftrightarrow \\ \frac{1}{g(x)} z^T x x^T z &\geq 0 \Leftrightarrow \\ \frac{1}{g(x)} (x^T z)^T (x^T z) = \frac{1}{g(x)} \| x^T z \|_2^2 &\geq 0 \end{align*} $$ which is true for all $z \in \mathbb{R}^n$ by the nonnegativity of the norm and of the fact that $g(x) > 0, \forall x$.