I hope it is okay that I re-post my question from Math.SE. I know it is very specific. I would be grateful for thoughts on how to tackle this problem. Any idea is welcome!

For $n \in \mathbb{N}$, let the positive real numbers $x_1, \dots, x_n, y_1, \dots, y_n$ satisfy the following interlacing inequalities: $$ 0 < x_1 < y_1 < x_2 < y_2 < \dots < x_n < y_n. $$ Show that $$ a_{ij} := \sum_{k=1}^n \frac{y_k^2 - x_k^2}{x_k (y_i^2 - x_k^2) (y_j^2 - x_k^2)}\Big( \prod_{l \neq k} \frac{y_l^2 - x_k^2}{x_l^2 - x_k^2} \Big) > 0 $$ for all $i,j = 1, \dots, n.$

My thoughts so far:

- The case $i=j$ is simple: $y_l^2 - x_k^2$ and $x_l^2-x_k^2$ have the same sign for all $k,l = 1, \dots, n.$ Since also $y_k^2 > x_k^2$ for all $k=1,\dots,n,$ every summand is positive.
- The problem is symmetric, so we only need to consider $i > j$. Then $(y_i^2 - x_k^2)(y_j^2-x_k^2) < 0$ if and only if $j < k \le i.$ I tried to apply Lagrange's interpolation formula, defining $$ p(x) := \frac12\prod_{l=1}^n (x_l-x) \prod_{k=1}^n (x_l+x) = \frac12\prod_{k=1}^n (x_l^2 - x^2). $$ This is a polynomial of degree $2n$, and it holds that $$ p'(x_k) = - x_k \prod_{l \neq k} (x_l^2-x_k^2) \qquad \text{ and } \qquad p'(-x_k) = x_k \prod_{l \neq k} (x_l^2 - x_k^2) $$ for $k = 1, \dots, n.$ Since $$ \frac1{(y_i^2-x_k^2)(y_j^2-x_k^2)} = \frac1{y_i^2-y_j^2} \Big( \frac1{y_i^2-x_k^2} - \frac1{y_j^2 - x_k^2}\Big), $$ we may write \begin{align*} a_{ij} &= \frac1{y_i^2 - y_j^2}\Big( \sum_{k=1}^n \frac{f_i(-x_k)}{p'(-x_k)(y_i+x_k)} - \sum_{k=1}^n \frac{f_j(-x_k)}{p'(-x_k)(y_j+x_k)}\Big)\\ &= \frac1{y_i^2 - y_j^2}\Big( \sum_{k=1}^n \frac{f_i(x_k)}{p'(x_k)(y_i-x_k)} - \sum_{k=1}^n \frac{f_j(x_k)}{p'(x_k)(y_j-x_k)}\Big) \end{align*} with $$ f_m(x) := -\frac{\text{sgn}(x)}{y_m+x} \prod_{l=1}^n (y_l^2-x^2), \qquad m \in \{i,j\}. $$ Unfortunately, this is not a polynomial.
- With a similar approach, one can show that $$ \sum_{k=1}^n \Big( \prod_{l \neq k} \frac{y_l^2 - x_k^2}{x_l^2 - x_k^2} \Big) \frac{y_k^2 - x_k^2}{(y_i^2 - x_k^2) (y_j^2 - x_k^2)} = \delta_{i=j}. $$ Maybe this is helpful?