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!