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!



How to prove positivity of determinant for these matrices?
At first, we prove that the determinant is non-zero, in other words, the matrix is non-singular. Assume the contrary, then by the linear dependency of the columns there exist real numbers $\lambda\_1,\dots,\lambda\_n$, not all equal to 0, such that $F(x\_i):=\sum\_j \frac{\lambda\_j}{g(x\_i-b\_j)}=0$ for all $i=1,2,\dots,n$. But the equation $F(x)=0$ is a polynomial equation with respect to $e^{2x}$ and the degree of a polynomial is less than $n$. So, it can not have $n$ distinct roots.


Now we note that the matrix is close to an identity when $x\_i=b\_i$ and $b\_i$'s are very much distant from each other, and the phase space of parameters $\{(x\_1,\dots,x\_n,b\_1,\dots,b\_n):x\_1<\dots<x\_n,b\_1<b\_2<\dots <b\_n\}$ is connected. Thus the sign of the determinant is always plus.


