Is this determinant always non-negative? For any $(a_1,a_2,\cdots,a_n)\in\mathbb{R}^n$, a matrix $A$ is defined by
$$A_{ij}=\frac1{1+|a_i-a_j|}$$
Is $\det(A)$ always non-negative? I did some numerical test and it seems to be true, but I have no idea how to prove it. Thanks!
 A: This answer is a suggestion as to how you may prove the result, but does not contain a proof.  Instead, I'll show how a similar problem has been answered, and offer pointers to claims that would show your result:
Consider the matrix:
$$A_{ij} = \frac{1}{1 + |a_i - a_j|^2}$$
Ultimately we will find that $\det(A)$ is always non-negative.  The reason is that $A$ itself is a positive definite matrix.  It is in the family of the Rational Quadratic Covariance Function: (aka Student T Kernel)

The rational quadratic covariance between two points separated by d distance units is given by
      $$C_{RQ}(d) = \Bigg(1+\frac{d^2}{2\alpha k^2}\Bigg)^{-\alpha}$$ 
  In this case $\alpha = 1$, $k = \frac{\sqrt{2}}{2}$.



(From Wikipedia): The Covariance Function of a space process $Z(x)$ is defined by
  $$C(x, y) = \textrm{Cov}(Z(x), Z(y))$$
  For locations $x_1, x_2, \ldots, x_N \in D$ the variance of every linear  combination:
      $$X=\sum_{i=1}^N w_i Z(x_i)$$
  can be computed as
      $$\operatorname{var}(X)=\sum_{i=1}^N \sum_{j=1}^N w_i C(x_i,x_j) w_j$$
  A function is a valid covariance function if and only if this variance is non-negative for all possible choices of $N$ and weights $w_1, \ldots, w_N$. A function with this property is called positive definite.

For this particular problem, the $x_i$ in the statement are the $a_i$ in the matrix $A$ defined above, and $C(x_i, x_j) = C_{RQ}(|a_i - a_j|)$.  It is well known (see below) that $C_{RQ}$ is positive definite as defined above and so for every linear combination:
$$\sum_{i=1}^N \sum_{j=1}^N w_i C_{RQ}(|a_i - a_j|) w_j \geq 0$$
i.e. $A$ is positive definite.  
Some references for the above (including the statement that $C_{RQ}$ is positive definite) can be seen at:
Gaussian Processes for Machine Learning Rasmussen, Williams 2006 and
Classes of Kernels for Machine Learning: A Statistics Perspective Genton 2001.
Now the kernel corresponding to the matrix you are interested in has been referred to as the  Generalized T Student Kernel.
$$C(x, y) = \frac{1}{1 + |x - y|^d}$$
At that blog, the following "proof" is offered:

The Generalized T-Student Kernel has been proven to be a Mercel Kernel, thus having a positive semi-definite Kernel matrix (Boughorbel, 2004).

The paper referenced is Conditionally Positive Definite Kernels For SVM Based Image Recognition - Boughorbel, Tarel, Boujemaa
If this proves to be true, then the determinant of your matrix is indeed non-negative.
A: We shall prove by mathematical induction that $A$ is positive semidefinite. The base case is trivial. Note that the positive semidefiniteness is preserved if we permute $a_1,\cdots,a_n$ or add the same constant to every $a_i$. So, we may assume that $a_1\ge a_2\ge \cdots\ge a_n=1$. Then $A$ becomes
$$
A=\left( \begin{array}{cccc} 
1 & \dfrac1{1+a_1-a_2} & \cdots & \dfrac1{a_1} \\
\dfrac1{1+a_1-a_2} & 1 & \cdots & \dfrac1{a_2} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac1{a_1} & \dfrac1{a_2} & \cdots & 1
\end{array}\right)
=\left(\frac1{1+a_{\min(i,j)}-a_{\max(i,j)}}\right)_{i,j=1,2,\ldots,n}
$$
Since $a_{nn}=1>0$, $A$ is PSD iff the Schur complement $S$ of $a_{nn}$ in $A$ in PSD. The Schur complement $S$ is given by
\begin{align*}
s_{ij}
&=\frac1{1+a_{\min(i,j)}-a_{\max(i,j)}}-\frac1{a_ia_j}\\
&=\frac1{1+a_{\min(i,j)}-a_{\max(i,j)}}-\frac1{a_{\min(i,j)}a_{\max(i,j)}}\\
&=\frac1{1+a_{\min(i,j)}-a_{\max(i,j)}}\cdot\frac{a_{\min(i,j)}+1}{a_{\min(i,j)}}\cdot\frac{a_{\max(i,j)}-1}{a_{\max(i,j)}}\\
&=\left(\frac1{1+a_{\min(i,j)}-a_{\max(i,j)}}\right)
\left(1-\frac1{a_{\max(i,j)}}\right)
\left(1+\frac1{a_{\min(i,j)}}\right).
\end{align*}
Therefore $S$ is a Hadamard product of three matrices. By induction assumption, the first matrix is PSD. It's easy to verify that the other two are PSD as well. Let $E_k$ denotes the matrix whose $k\times k$ leading principal submatrix contains only ones and all other entries are zero. Since $1-\frac1{a_i}$ is a decreasing sequence, the matrix $\left(1-\frac1{a_{\max(i,j)}}\right)_{i,j=1,2,\ldots,n-1}$ is a nonnegative weighted sum of $E_1,E_2,\ldots,E_{n-1}$. Similarly, as $1+\frac1{a_i}$ is increasing, $\left(1+\frac1{a_{\min(i,j)}}\right)_{i,j=1,2,\ldots,n-1}$ is also a nonnegative weighted sum of PSD matrices (but whose trailing principal submatrices, rather than the leading ones, contain ones). So, by Schur product theorem, $S$ is PSD and hence $A$ is PSD.
A: The function $f:x\mapsto\frac1{1+x}$ is non-constant and completely monotone on $[0,\infty)$  (i.e. all even-order derivatives of $f$ are nonnegative and all odd-order derivatives of $f$ are nonpositive). Therefore, by Schoenberg's interpolation theorem, the matrix $\left(f(\|a_i-a_j\|_2)\right)_{i,j=1,2,\ldots,n}$ is positive semidefinite, and it is positive definite iff all $a_i$s are distinct.
