I hope it is okay that I re-post [my question](https://math.stackexchange.com/questions/2156119) 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?



Inequality connected to Lagrange's interpolation formula
Write $x\_i$ for $x\_i^2$ and $y\_i$ for $y\_i^2$, our sum is
$$
a\_{ij}=\sum\_{k=1}^n \frac{f(x\_k)}{\prod\_{l\ne k} (x\_k-x\_l)},\,f(t)=-t^{-1/2}\prod\_{l\ne i,j} (t-y\_j).
$$  

By Lagrange interpolation, it is a coefficient of $t^{n-1}$ in the polynomial $g(t)$ of degree at most $n-1$, which interpolates the function $f$ in the points $x\_1,\dots,x\_n$. Thus by Rolle's theorem we have $a\_{ij}\cdot (n-1)!=f^{(n-1)}(\theta)$ for a certain positive $\theta$. But $(n-1)$-st derivative of $f$ is clearly positive term-wise: $f=(-1)^{n-1}(c\_1t^{-1/2}-c\_2t^{1/2}+c\_3t^{3/2}-\dots+(-1)^{n-2}c\_{n-1}t^{n-5/2})$ for positive coefficients $c\_1,c\_2,\dots,c\_{n-1}$.


Note that we did not use that $x$'s and $y$'s interlace, only that they are positive.


