This question is a follow-up on [another MO query here](https://mathoverflow.net/questions/263265/inequality-connected-to-lagranges-interpolation-formula).



> 
> **Question.** For $r\geq$ an integer, is it true that there exists homogeneous symmetric polynomial $P\_r(x\_1,\dots,x\_n)$ with **positive coefficients** such that
>  $$(-1)^r\sum\_{k=1}^nx\_k^{2r-1}\prod\_{\ell\neq k}^{1,n}\frac1{x\_{\ell}^2-x\_k^2}
> =\frac{P\_r(x\_1,\dots,x\_n)}{\prod\_{j=1}^nx\_j}\prod\_{i<j}^{1.n}\frac1{x\_i+x\_j} \,\, ?$$
>  I would be content with $0\leq r\leq n$. Is there an **explicit** formulation of $P\_r$?
> 
> 
> 


**Remark.** The factor $\prod\_jx\_j$, in the denominator of the RHS, is needed only when $r=0$.


For example, when $n=3, r=0$, the RHS is $\frac{x\_1+x\_2+x\_3}{x\_1x\_2x\_3(x\_1+x\_2)(x\_1+x\_3)(x\_2+x\_3)}$. 


If $n=4, r=2$ then the RHS takes the form
$$\frac{x\_1x\_2x\_3+x\_1x\_2x\_4+x\_1x\_3x\_4+x\_2x\_3x\_4}{(x\_1+x\_2)(x\_1+x\_3)(x\_1+x\_4)(x\_2+x\_3)(x\_2+x\_4)(x\_3+x\_4)}.$$



Lagrange interpolation vs homogeneous symmetric polynomials?
Denote by $\delta\_n$ the partition $(n-1,n-2,\dots,1)$ of $\binom{n}2$, and the associated schur function is
$$s\_{\delta\_n}(x\_1,\dots,x\_n)=\prod\_{i<j}(x\_i+x\_j).$$
Consider now the ratio of determinants
$$\begin{vmatrix}
x\_1^{2r-1} & x\_2^{2r-1} &\cdots & x\_n^{2r-1} \\
x\_1^{2n-4} & x\_2^{2n-4} &\cdots & x\_n^{2n-4} \\
\vdots &\vdots & \ddots & \vdots\\
x\_1^2 & x\_2^2 & \cdots & x\_n^2 \\
1 & 1& \cdots &1
\end{vmatrix} \cdot \begin{vmatrix}
x\_1^{2n-2} & x\_2^{2n-2} &\cdots & x\_n^{2n-2} \\
x\_1^{2n-4} & x\_2^{2n-4} &\cdots & x\_n^{2n-4} \\
\vdots &\vdots & \ddots & \vdots\\
x\_1^2 & x\_2^2 & \cdots & x\_n^2 \\
1 & 1& \cdots &1
\end{vmatrix}^{-1}$$
This evaluates to $\sum\_{k=1}^nx\_k^{2r-1}\prod\_{\ell\neq k}^{1,n}\frac1{x\_{\ell}^2-x\_k^2}$. Proof: Expand the first determinant by the first row, and use the fact that the second is just a Vandermonde that evaluates to $\prod\_{i<j} (x\_j^2-x\_i^2)$.


Therefore your polynomial $P\_r(x\_1,x\_2,\dots,x\_n)$ is equal to
$$(-1)^r \begin{vmatrix}
x\_1^{2r} & x\_2^{2r} &\cdots & x\_n^{2r} \\
x\_1^{2n-3} & x\_2^{2n-3} &\cdots & x\_n^{2n-3} \\
\vdots &\vdots & \ddots & \vdots\\
x\_1^3 & x\_2^3 & \cdots & x\_n^3 \\
x\_1 & x\_2& \cdots &x\_n
\end{vmatrix} \cdot \begin{vmatrix}
x\_1^{2n-2} & x\_2^{2n-2} &\cdots & x\_n^{2n-2} \\
x\_1^{2n-4} & x\_2^{2n-4} &\cdots & x\_n^{2n-4} \\
\vdots &\vdots & \ddots & \vdots\\
x\_1^2 & x\_2^2 & \cdots & x\_n^2 \\
1 & 1& \cdots &1
\end{vmatrix}^{-1}\cdot s\_{\delta\_n}(x\_1,x\_2,\dots,x\_n)$$
When $0\le r\le n$ we can use the definition of Schur polynomials as ratios of determinants to simplify this expression as
$$s\_{\lambda\_{r,n}}(x\_1,x\_2,\dots,x\_n)$$
Where the partition $\lambda\_{r,n}$ is given by $(n-2,n-3,\dots,r+1,r,r,r,r-1,\dots,2,1)$. Therefore, not only is $P\_{r}(x\_1,x\_2,\dots,x\_n)$ symmetric with positive integer coefficients, but we also get a combinatorial interpretation of these coefficients as counting semistandard tableaux of shape $\lambda\_{r,n}$.


