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let $a_{1},a_{2},\cdots,a_{n+1}$ be a sequence of distinct non-zero real numbers with $$\sum_{j=1}^{n+1}a^2_{j}=1,~~~\sum_{j=1}^{n+1}a_{j}=0$$ show this $$0<\sum_{k=1}^{n+1}\dfrac{1}{|a_{k}|}\prod_{j=1,j\neq k}^{n+1}\dfrac{a_{k}}{a_{k}-a_{j}}\le\sqrt{2}\tag{1}$$

I found the equality on the right-hand side when $n=1$.But I can't prove this inequality $(1)$.First of all, this inequality is a bit like Lagrange's interpolation formula Prove $1 + \sum_{i=0}^n(\frac1{x_i}\prod_{j\neq i}(1+\frac1{x_j-x_i}))=\prod_{i=0}^n(1+\frac1{x_i})$, I'm going to try to prove it using Lagrange's interpolation formula but can't it

math110
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    Isn't Lagrange interpolation formula for polynomials? – Divide1918 Dec 31 '20 at 06:54
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    $\sum_{k=1}^{n}\dfrac 1{|a_k|}\prod_{j\ne k,j\le n} \dfrac{a_k}{a_k-a_j}\le \sqrt 2$

    $\sum_{k=1}^{n+1}\dfrac 1{|a_k|}\prod_{j\ne k,j\le n+1} \dfrac{a_k}{a_k-a_j}=\sum_{k=1}^{n}\dfrac 1{|a_k|(a_k-a_{n+1})}\prod_{j\ne k,j\le n} \dfrac{a_k}{a_k-a_j}+\dfrac 1{|a_{n+1}|}\prod_{j\ne n+1}\dfrac{a_{n+1}}{a_{n+1}-a_j}$.

    The first part is $\le \dfrac {\sqrt2}{a_l-a_{n+1}}$ whence $l$ is such that $a_l-a_{n+1}$ is minimal among all $1\le l\le n$.

     – Divide1918 Dec 31 '20 at 07:24
  • Now posted on MO, too: How prove this Webb inequality? – Martin Sleziak Jan 14 '21 at 04:29

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