TITLE: How prove this Webb inequality?
QUESTION [7 upvotes]: I posted this question 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.

REPLY [4 votes]: 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.