I posted this [question](https://math.stackexchange.com/questions/3967715/prove-0-sum-k-1n1-frac1a-k-prod-j-1-j-neq-kn1-fraca-k) on Math StackExchange but did not get any answer. I am trying my luck here.



> 
> Let $a\_{1},a\_{2},\dotsc,a\_{n+1}$ be a sequence of distinct non-zero real numbers with
> $$\sum\_{j=1}^{n+1}a^2\_{j}=1,\qquad\sum\_{j=1}^{n+1}a\_{j}=0.$$
> Show
> \begin{equation}
> \tag{1}
> \label{1}
> 0<\sum\_{k=1}^{n+1}\dfrac{1}{\lvert a\_{k}\rvert}\prod\_{\substack{j=1 \\ j\neq k}}^{n+1}\dfrac{a\_{k}}{a\_{k}-a\_{j}}\le\sqrt{2}.
> \end{equation}
> 
> 
> 


I found the equality on the right-hand side when $n=1$. But I can't prove this inequality \eqref{1}. First of all, this inequality is a bit like Lagrange's interpolation formula <https://math.stackexchange.com/questions/500139/prove-1-sum-i-0n-frac1x-i-prod-j-neq-i1-frac1x-j-x-i-prod-i>. I tried to prove it using Lagrange's interpolation formula but can't.



How prove this Webb inequality?
Here is the proof that the sum is non-negative.


Denote $f(x)=x^{n-2}|x|=x^n/|x|$. Then
$$
A:=\sum\_{k=1}^{n+1}\dfrac{1}{|a\_{k}|}\prod\_{j=1,j\neq k}^{n+1}\dfrac{a\_{k}}{a\_{k}-a\_{j}}=
[x^n]\sum\_{k=1}^{n+1}f(a\_k)\prod\_{j\ne k}\frac{x-a\_j}{a\_k-a\_j}=:[x^n]h(x),
$$
where the polynomial $h(x)=Ax^n+\ldots$, $\deg h\leqslant n$, interpolates $f$ in points $a\_1,\ldots,a\_{n+1}$.


The function $h(x)-f(x)$ is $(n-2)$ times continuously differentiable and has $n+1$ roots at $a\_i$'s, thus by Rolle theorem its $(n-2)$-st derivative $\frac{n!}2Ax^2+Bx+C-(n-1)!|x|$ has three distinct roots. But if $A<0$, this function is strictly concave and can not have three roots. So $A\geqslant 0$.


Also this sum appears in the theory of splines and in the algebraic combinatorics / integrable probability. Probably everything is done and written somewhere.


