It's a Theorem about polynomials proved by Schur in 1918

Beautiful problem Do you have reference for proof of Weida’s theorem ?

Any proof to this beauty

Moubinool wrote: Do you have reference for proof of Weida’s theorem ? I think he means Vieta's Theorem...

No FloorX not Vieta’s theorem see the proof of Mr.Trick it is written « By Weida’s theorem»

Moubinool wrote: No FloorX not Vieta’s theorem see the proof of Mr.Trick it is written « By Weida’s theorem» So it's a typo.

Moubinool wrote: Beautiful problem Do you have reference for proof of Weida’s theorem ? Sorry,it's "Vieta's theorem".(My English sucks)

One solution is to suppose some points A_1~A_2n-2 on the unit circle,for 1≤k≤n, arg(A_k)=arccos(x_k) and arg(A_2n-k)=-arccos(x_k) Obviously we can suppose x_1=1 and x_n=-1 Use cosx-cosy=2sin[(x+y)/2]*sin[(x-y)/2] ,transform the problem :For any epsilon >0, and n is big enough , the product of |A_iA_j| (1≤i<j≤2n, i+j≠2n)<(1+epsilon)^(0.5*n^2). f(x)=ln(cos x), f"(x)=-1/(sin x)^2 < 0.Use Jensen Inequality again and agdin,add some details and discussion,the problem will be solved

Any reference about the theorems Mr.Trick used

Any other elementary solution?

The Schur's result referenced in $\#2$ is about monic polynomial with integer coefficients. In our case it's not given that $x_1,x_2,\dots,x_n$ are roots of polynomial like that. That is, we have more freedom to vary $x_1,x_2,\dots,x_n$ and could expect $C$ be grater than $1/2$. Nevertheless, the result apparently is the same $C=1/2$. Note that $\prod_{1\le i<j\le n}(x_i-x_j)$ is the Vandermonde's determinant $V_n(x_1,\dots,x_n)$ so, the question boils down to find (estimate) the maximum $M_n$ of $V_n$ when $x_i\in[-1,1]$. This question appears to be well known. The knots $x_i$ where this max takes place are called Fekete points and are the roots of $(1-x^2)P'(x)$ where $P(x)$ is the Legendre polynomial of degree $n-1$. See also this entry in mathoverflow. Let $c_n:=M_n^{1/\binom{n}{2}}$. Then $C=\inf\{c_n, n\in\mathbb{N}\}$. In a paper from 1923 (referenced in the wiki page on Fekete problem) M. Fekete delved into this problem. Unfortunately the paper is in German, which I don't understand. For instance, $c_n$ is monotone decreasing. I don't think this problem has a short and elementary solution. Not that it's too high-tech, but at least some preliminary knowledge is required.

Horrible problem by 王彬

Mr.Trick wrote: It's a Theorem about polynomials proved by Schur in 1918 Can anyone explains the limit??

Please would anyone explain the limit??

The official solution is huge amount of calculation after $x_i=\cos\theta_j,$ here is an elegant proof. For all positive integer $n,$ let $T_n(x)$ be the $n$ th Chebyshev polynomial, here define $T_0(x):=1.$ Then \begin{align*} \prod_{1\le i<j\le n}(x_i-x_j) &=\left|\left|\begin{matrix} 1&1&\cdots&1\\ x_1&x_2&\cdots &x_n\\ \vdots&\vdots&\ddots&\vdots\\ x_1^{n-1}&x_2^{n-1}&\cdots &x_n^{n-1} \end{matrix}\right|\right|\\ &=\left|\left|\begin{matrix} 1&1&\cdots&1\\ T_1(x_1)/2&T_1(x_2)/2&\cdots &T_1(x_n)/2\\ \vdots&\vdots&\ddots&\vdots\\ T_{n-1}(x_1)/2^{n-2}&T_{n-1}(x_2)/2^{n-2}&\cdots &T_{n-1}(x_n)/2^{n-2} \end{matrix}\right|\right|\\ &=\dfrac 1{2^{\frac{(n-1)(n-2)}2}}\left|\left|\begin{matrix} 1&1&\cdots&1\\ T_1(x_1)&T_1(x_2)&\cdots &T_1(x_n)\\ \vdots&\vdots&\ddots&\vdots\\ T_{n-1}(x_1)&T_{n-1}(x_2)&\cdots &T_{n-1}(x_n) \end{matrix}\right|\right|\\ &\le\dfrac 1{2^{\frac{(n-1)(n-2)}2}}\prod_{i=1}^n\sqrt{\sum_{j=0}^{n-1}T_j^2(x_i)}\le\dfrac {\sqrt n^n}{2^{\frac{(n-1)(n-2)}2}}. \end{align*}This gives $C\le\frac 12.$ The full solution can be found in attached file.

Thanks for sharing @EthanWYX2009