TITLE: How to prove positivity of determinant for these matrices?
QUESTION [19 upvotes]: Let $g(x) = e^x + e^{-x}$. For $x_1 < x_2 < \dots < x_n$ and $b_1 < b_2 < \dots < b_n$, I'd like to show that the determinant of the following matrix is positive, regardless of $n$:
$\det \left (\begin{bmatrix} 
\frac{1}{g(x_1-b_1)} & \frac{1}{g(x_1-b_2)} & \cdots & \frac{1}{g(x_1-b_n)}\\
\frac{1}{g(x_2-b_1)} & \frac{1}{g(x_2-b_2)} & \cdots & \frac{1}{g(x_2-b_n)}\\
\vdots & \vdots & \ddots & \vdots \\
\frac{1}{g(x_n-b_1)} & \frac{1}{g(x_n-b_2)} & \cdots & \frac{1}{g(x_n-b_n)}
\end{bmatrix} \right ) > 0$.
Case $n = 2$ was proven by observing that $g(x)g(y) = g(x+y)+g(x-y)$, 
and
$g(x_2 - b_1)g(x_1-b_2) = g(x_1+x_2 - b_1-b_2)+g(x_2-x_1+b_2-b_1) > g(x_1+x_2 - b_1-b_2)+g(x_2-x_1-b_2+b_1) = g(x_1-b_1)g(x_2-b_2)$
However, things get difficult for $n \geq 3$. Any ideas or tips?
Thanks!

REPLY [25 votes]: To complement Fedor's answer, here is more explicit proof.
Let the original matrix be $G$. Let $D_x :=\text{Diag}(e^{x_1},\ldots,e^{x_n})$. Then, we can write
\begin{equation*}
G = D_x C D_b,\quad\text{where}\ C = \left[ \frac{1}{e^{2x_i}+e^{2b_j}}\right]_{i,j=1}^n.
\end{equation*}
To prove that $\det(G)>0$ it thus suffices to prove that $\det(C)>0$. Notice now that $C$ is nothing but a Cauchy matrix, and by explicitly writing its determinant out (under the hypotheses on $x$ and $b$) we can easily conclude that $\det(C)>0$.

Remark. The above argument actually proves that $\text{sech}(x-y)$ is a Totally positive kernel (because the $k(x,y) := 1/(x+y)$ is known to be a TP kernel).