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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?
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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.

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  • $\begingroup$ Why is it clear that $f^{(n-1)}$ is positive? $\endgroup$ – Elias Strehle Feb 28 '17 at 9:53
  • $\begingroup$ look at the formula for $f$ (written now in the answer) and differentiate each term $n-1$ times. The sign is always positive. $\endgroup$ – Fedor Petrov Feb 28 '17 at 10:46
  • $\begingroup$ I am having trouble with the signs. The Lagrange polynomial is $$\Big(\sum_{k=1}^n \frac{f(x_k)}{\prod_{l \neq k}(x_l-x_k)}\Big)x^{n-1} + ...$$ (not $(x_k - x_l)$). If I follow your proof I get that $(-1)^{n-1}a_{ij} > 0$? $\endgroup$ – Elias Strehle Feb 28 '17 at 12:53
  • $\begingroup$ Why? $x_k-x_l$, not $x_l-x_k$ $\endgroup$ – Fedor Petrov Feb 28 '17 at 13:19
  • $\begingroup$ To interpolate $f$ at the $x_1, \dots, x_n$, I chose $$ g(x) = \sum_{k=1}^n f(x_k) \Big( \prod_{l \neq k} \frac{x-x_l}{x_k-x_l}\Big) = \Big(\sum_{k=1}^n \frac{f(x_k)}{\prod_{l\neq k} x_k-x_l} \Big) x^{n-1} + \tilde{g}(x), $$ for some polynomial $\tilde{g}$ of degree $n-2$. $\endgroup$ – Elias Strehle Feb 28 '17 at 13:39
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The expression is equivalent to $$ a_{ij}=\sum_k \frac{\prod_{l\neq i,j} (y_l^2-x_k^2) }{x_k\prod_{l\neq k} (x_l^2-x_k^2)}. $$ This is a linear function in each $y^2_m$ for every $m$. Assume that there exist $y_m\in [x_1,x_n]$. Then the term gets smaller or equal if we replace $y_m$ by $x_1$ or $x_n$. However after the cancellations what is left is an expression of the same form with a smaller $n$. Hence one can prove the inequality by induction.

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  • $\begingroup$ The assumption $y_m \in [x_1,x_n]$ makes this difficult: We are cancelling $y_m$ and $x_1$ (or $x_n$). For the induction to work, we now need another $y_{m'} \in [x_2, x_n]$ etc. $\endgroup$ – Elias Strehle Feb 28 '17 at 11:27
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Just assume $x_k, y_k>0$ without interlacing. If necessary by indexing, write $$a=\sum_k \frac{\prod_{l=1}^{n-2}(y_l^2-x_k^2)}{x_k\prod_{l\neq k}^{1,n}(x_l^2-x_k^2)}.$$ If $e_m(y_1^2,\dots,y_{n-2}^2)$ denotes the elementary symmetric polynomials, then $$\prod_{l=1}^{n-2}(y_l^2-x_k^2)=\sum_{r=0}^{n-2}(-1)^rx_k^{2r}\,e_{n-2-r}(y_1^2,\dots,y_{n-2}^2).$$ Since $e_m>0$, to prove $a>0$ it suffices to check that $$(-1)^r\sum_{k=1}^nx_k^{2r-1}\prod_{\ell\neq k}^{1,n}\frac1{x_{\ell}^2-x_k^2}>0 \qquad \text{for each $\,\,0\leq r\leq n-2$}.$$ I claim that there exists a homogeneous symmetric polynomial $P_r(x_1,\dots,x_n)$ with positive coefficients such that $$(-1)^r\sum_{k=1}^nx_k^{2r-1}\prod_{\ell\neq k}^{1,n}\frac1{x_{\ell}^2-x_k^2} =\frac{P_r(x_1,\dots,x_n)}{\prod_{j=1}^nx_j}\prod_{i<j}^{1.n}\frac1{x_i+x_j}.$$ This has now become a new query on MO. Update. There's a cute answer given now. Check it out!

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  • $\begingroup$ Maybe Fedor Petrov's answer helps with your query as well? I can post my full proof if you want. $\endgroup$ – Elias Strehle Feb 28 '17 at 18:39

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