Problem 2518

For sufficiently large $n$, we guess the optimal choice of $(x_1, \dots, x_n)$ on $[-1, 1]$ is close to the projection of points uniformly distributed on the unit circle onto the $x$ axis. To this end, we introduce the reference choice
$$a_k = \cos \vartheta_k = \cos \frac{2k-1}{2n} \pi, \quad k = 1, 2, \dots, n.$$
For $A = \{a_1, a_2, \dots, a_n\}$, the product
$$P_A = \prod_{1 \le i < j \le n} (a_i - a_j)$$
and
$$\sum_{i \neq k} \frac{1}{a_k - a_i}, \quad k = 1, \dots, n$$
will be used later.

Consider the $n$th degree Chebyshev polynomial of the first kind $T(X) = T_n(X)$, with leading coefficient $2^{n-1}$ and $T(\cos \vartheta) = \cos n\vartheta$.
Substitute $X = \cos \vartheta$, and differentiate to find $T'(\cos \vartheta)$, $T''(\cos \vartheta)$:
$$\begin{align*} T'(\cos \vartheta) &= \frac{d \cos n\vartheta}{d \cos \vartheta} = \frac{-n \sin n\vartheta \times d\vartheta}{-\sin \vartheta \times d\vartheta} = n \times \frac{\sin n\vartheta}{\sin \vartheta}; \ T''(\cos \vartheta) &= \frac{dT'(\cos \vartheta)}{d \cos \vartheta} = n \times \frac{d \frac{\sin n\vartheta}{\sin \vartheta}}{d \cos \vartheta} \ &= n \times \frac{\frac{n \cos n\vartheta}{\sin \vartheta} \times d\vartheta - \frac{\cos \vartheta \sin n\vartheta}{\sin^2 \vartheta \times d\vartheta}}{-\sin \vartheta \times d\vartheta} \ &= n \times \frac{\cos \vartheta \sin n\vartheta - n \times \cos n\vartheta \sin \vartheta}{\sin^3 \vartheta}. \end{align*}$$
For $k = 1, 2, \dots, n$, denote $\vartheta_k = \frac{2k-1}{2n}\pi$. We have
$$T(a_k) = T(\cos \vartheta_k) = \cos(n\vartheta_k) = 0,$$
and hence all the roots of $T(X)$ are $A = \{a_1, a_2, \dots, a_n\}$. Define
$$H(X) = \frac{1}{2^{n-1}} T(X) = (X - a_1)(X - a_2) \cdots (X - a_n).$$
From another perspective, the first derivative $H'(X) = \frac{dH(X)}{dX}$ is the sum of all products of $n-1$ terms in
$$B = \{X - a_1, X - a_2, \dots, X - a_n\}$$
(there are $\binom{n}{n-1}$ products), denoted as $H'(X) = \sigma_{n-1}(B)$. Furthermore, the second derivative $H''(X) = \frac{dH'(X)}{dX}$ is twice the sum of all products of $n-2$ terms in $B$ (there are $\binom{n}{n-2}$ products), denoted as $H''(X) = 2\sigma_{n-2}(B)$.

When $X = a_k \in A$, we have $X - a_k = 0$ in $B$, and
$$H'(a_k) = \sigma_{n-1}(B) = \prod_{i \neq k} (a_k - a_i),$$
$$H''(a_k) = 2\sigma_{n-2}(B) = 2 \prod_{i \neq k} (a_k - a_i) \sum_{i \neq k} \frac{1}{a_k-a_i}.$$
Substitute $X = a_k = \cos \vartheta_k \in A$ in $T'(X)$ and $T''(X)$ to obtain
$$T'(a_k) = T'(\cos \vartheta_k) = n \times \frac{(-1)^{k-1}}{\sin \vartheta_k},$$
$$T''(a_k) = T''(\cos \vartheta_k) = n \times \frac{(-1)^{k-1} \cos \vartheta_k}{\sin^3 \vartheta_k}.$$
Therefore,
$$\sum_{i \neq k} \frac{1}{a_k - a_i} = \frac{H''(a_k)}{2H'(a_k)} = \frac{T''(a_k)}{2T'(a_k)} = \frac{\cos \vartheta_k}{2 \sin^2 \vartheta_k}, \quad k = 1, 2, \dots, n.$$
Moreover,
$$\begin{align*} P_A^2 &= (-1)^{\frac{n(n-1)}{2}} \prod_{k=1}^n \prod_{i \neq k} (a_k - a_i) \ &= (-1)^{\frac{n(n-1)}{2}} \prod_{k=1}^n H'(\cos \vartheta_k) \ &= \frac{n^n}{2^{n(n-1)} \prod_{k=1}^n \sin \vartheta_k}. \end{align*}$$
Notice that the complex roots of $(z+1)^n + 1 = 0$ are $\{e^{\frac{2k-1}{n}\pi i} - 1, k = 1, \dots, n\}$ with magnitudes $2 \sin \frac{2k-1}{2n}\pi = 2 \sin \vartheta_k$. The product of the magnitudes is $\prod_{k=1}^n 2 \sin \vartheta_k = 2$, implying
$$P_A^2 = \frac{n^n}{2^{n(n-1)} \prod_{k=1}^n \sin \vartheta_k} = \frac{n^n}{2^{n(n-1)} 2^{-(n-1)}} \Rightarrow P_A = \left(\frac{1}{2}\right)^{\frac{n(n-1)}{2}} n^{\frac{n}{2}} 2^{\frac{n-1}{2}},$$
and hence $C = \frac{1}{2}$ satisfies the problem statement. In the following, we prove $C \le \frac{1}{2}$.
For any positive integer $n$, assume $1 \ge x_1 > x_2 > \cdots > x_n \ge -1$ and
$$P^* = \prod_{1 \le i < j \le n} (x_i - x_j) \ge C^{\frac{n(n-1)}{2}}.$$
We make a comparison with the reference choice:
$$\begin{align*} F &= \sum_{1 \le i < j \le n} \frac{x_i - x_j}{a_i - a_j} \ &= \sum_{k=1}^n \left( x_k \times \sum_{i \neq k} \frac{1}{a_k - a_j} \right) \ &= \sum_{k=1}^n \left( x_k \times \frac{\cos \vartheta_k}{2 \sin^2 \vartheta_k} \right) \ &\le \frac{1}{2} \sum_{k=1}^n \frac{|\cos \vartheta_k|}{\sin^2 \vartheta_k} \end{align*}$$
Evaluating
$$\begin{align*} G &= \sum_{k=1}^n \cot^2 \vartheta_k = \sum_{k=1}^n \cot^2 \frac{2k-1}{2n} \pi \ &= \sum_{l=1}^{2n-1} \cot^2 \frac{l\pi}{2n} - \sum_{l=1}^{n-1} \cot^2 \frac{l\pi}{n}, \end{align*}$$
using the well-known identity
$$\sum_{k=1}^{m-1} \cot^2 \frac{k\pi}{m} = \frac{(m-1)(m-2)}{3},$$
it follows that
$$G = \frac{(2n-1)(2n-2)}{3} - \frac{(n-1)(n-2)}{3} = n(n-1).$$
Now, by Cauchy's inequality,
$$\begin{align*} (2F)^2 &\le \left( \sum_{k=1}^n \frac{|\cos \vartheta_k|}{\sin^2 \vartheta_k} \right)^2 \\ &\le \sum_{k=1}^n \frac{\cos^2 \vartheta_k}{\sin^2 \vartheta_k} \times \sum_{k=1}^n \frac{1}{\sin^2 \vartheta_k} \\ &= G \times (G + n) \\ &= (n-1)n^3. \end{align*}$$
Finally, by AM-GM inequality, we arrive at
$$\begin{align*} \frac{P^*}{P_A} &= \prod_{1 \le i < j \le n} \frac{x_i - x_j}{a_i - a_j} \\ &\le \left( \frac{2}{n(n-1)} \sum_{1 \le i < j \le n} \frac{x_i - x_j}{a_i - a_j} \right)^{\frac{n(n-1)}{2}} \\ &< \left( \frac{n}{n-1} \right)^{\frac{n(n-1)}{4}} < e^{\frac{n}{4}}. \\ C^{\frac{n(n-1)}{2}} &\le P^* < e^{\frac{n}{4}} \times P_A = \left( \frac{1}{2} \right)^{\frac{n(n-1)}{2}} \times n^{\frac{n}{2}} 2^{\frac{n-1}{2}} e^{\frac{n}{4}}. \end{align*}$$
The above inequality holds for all positive integers $n$. By taking $n$ sufficiently large, we deduce that $C \le \frac{1}{2}$. $\square$

(modified from Wang Yichuan's solution)
First prove $C \le \frac{1}{2}$. Without loss of generality, assume
$1 \ge x_1 \ge \cdots \ge x_n \ge -1,$
where $x_i = \cos \vartheta_i$ ($1 \le i \le n$), $0 \le \vartheta_1 \le \cdots \le \vartheta_n \le \pi$. Let $p_j = e^{i\vartheta_j}$,
$q_j = e^{-i\vartheta_j}$ be complex numbers. Clearly, on the complex plane, $p_1, \dots, p_n,$ $q_n, \dots, q_1$ correspond to $2n$ points anticlockwise on the unit circle, which we denote in order as $A_1, \dots, A_{2n}$. It follows that
$$\prod_{1 \le j < k \le n} (x_j - x_k) = \prod_{1 \le j < k \le n} (\cos \vartheta_j - \cos \vartheta_k) = \prod_{1 \le j < k \le n} 2 \sin \frac{\vartheta_j + \vartheta_k}{2} \sin \frac{\vartheta_j - \vartheta_k}{2}. \quad \textcircled{1}$$
By the law of sine, for $1 \le j < k \le n$,
$$2 \sin \frac{\vartheta_j + \vartheta_k}{2} = |p_j - q_k| = |p_k - q_j|,$$
$$2 \sin \frac{\vartheta_j - \vartheta_k}{2} = |p_j - p_k| = |q_j - q_k|.$$
Hence,
$$2 \sin \frac{\vartheta_j + \vartheta_k}{2} \sin \frac{\vartheta_j - \vartheta_k}{2} = \frac{1}{2} \sqrt{|p_j - q_k| |p_k - q_j| |p_j - p_k| |q_j - q_k|}$$
$$= \frac{1}{2} \sqrt{|A_j A_{2n+1-k}| |A_k A_{2n+1-j}| |A_j A_k| |A_{2n+1-j} A_{2n+1-k}|}$$
Plug the above identity into (1), to obtain
$$\begin{aligned} & \prod_{1 \le j < k \le n} (x_j - x_k) \\ &= \prod_{1 \le j < k \le n} \frac{1}{2} \sqrt{|A_j A_{2n+1-k}| |A_k A_{2n+1-j}| |A_j A_k| |A_{2n+1-j} A_{2n+1-k}|} \\ &= \left(\frac{1}{2}\right)^{\frac{n(n-1)}{2}} \sqrt{\prod_{1 \le j < k \le 2n, j+k \ne 2n+1} |A_j A_k|}. \end{aligned} \quad \textcircled{2}$$
Now we estimate $\prod_{1 \le j < k \le 2n, j + k \ne 2n + 1} |A_j A_k|$. Divide $A_j A_k$ ($1 \le j < k \le 2n$) into $n$ groups
$\Omega_1 : A_1 A_2, A_2 A_3, \dots, A_{2n} A_1;$
$\Omega_2 : A_1 A_3, A_2 A_4, \dots, A_{2n} A_2;$
$\dots$
$\Omega_{n-1} : A_1 A_2, A_2 A_3, \dots, A_{2n} A_{n-1};$
$\Omega_n : A_1 A_{n+1}, A_2 A_{n+2}, \dots, A_n A_{2n}.$
Notice that for each $1 \le j \le n-1$, the sum of the inscribed angles subtended by all chords in $\Omega_j$ is $j\pi$; for each $\Omega_j$, let $\lambda_j$ be the number of chords $A_s A_t$ such that $s+t \ne 2n+1$, then it must be $\lambda_j = 2n$ or $\lambda_j = 2n-2$. From the law of sine and Jensen's inequality applied to the convex function $\ln(\sin x)$ ($x \in (0, \frac{\pi}{2})$), it follows that
$$\prod \{|A_s A_t| : A_s A_t \in \Omega_j, s + t \neq 2n + 1\} \leq \left(2 \sin \frac{j\pi}{\lambda_j}\right)^{\lambda_j}.$$
On one hand, due to $\sin \frac{j\pi}{\lambda_j} \le 1$,
$$\left(2 \sin \frac{j\pi}{\lambda_j}\right)^{\lambda_j} \le 2^{2n} \left(\sin \frac{j\pi}{\lambda_j}\right)^{2n-2} \le 2^{2n} \left(\sin \frac{j\pi}{2n-2}\right)^{2n-2}.$$
On the other hand, we have
$$\prod \{|A_s A_t| : A_s A_t \in \Omega_n\} \le 2^n.$$
Hence,
$$\begin{align*} \prod_{1 \le j < k \le 2n, j + k \ne 2n + 1} |A_j A_k| &\le 2^n \cdot \prod_{j=1}^{n-1} \left( 2^{2n} \left( \sin \frac{j\pi}{2n-2} \right)^{2n-2} \right) \\ &= 2^{3n-2} \left( \prod_{j=1}^{n-1} \left( 2 \sin \frac{j\pi}{2n-2} \right) \right)^{2n-2} \\ &\le 2^{3n-2} \cdot (2\sqrt{n})^{2n-2}, \end{align*}$$
and in the last step we used the trigonometric identity
$$\prod_{j=1}^m \left( 2 \sin \frac{j\pi}{2m} \right) = 2\sqrt{m}.$$
Now plug the inequality into (2), yielding
$$C^{\frac{n(n-1)}{2}} \le \prod_{1 \le j < k \le n} (x_j - x_k) \le \left(\frac{1}{2}\right)^{\frac{n(n-1)}{2}} \cdot 2^{\frac{3n-2}{2}} \cdot (2\sqrt{n})^{n-1}.$$
Letting $n \to \infty$, $C \le \frac{1}{2}$.
It remains to prove that $C = \frac{1}{2}$ can be attained. Let $(\vartheta_1, \dots, \vartheta_n) = \left(\frac{\pi}{2n}, \frac{3\pi}{2n}, \dots, \frac{(2n-1)\pi}{2n}\right)$. Evidently, the complex numbers $p_j = e^{i\vartheta_j}$, $q_j = e^{-i\vartheta_j}$ correspond to a regular $2n$-gon $A_1 \dots A_{2n}$ in the complex plane. From (2), we get
$$\prod_{1 \le j < k \le n} (x_j - x_k) = \left(\frac{1}{2}\right)^{\frac{n(n-1)}{2}} \sqrt{\prod_{\substack{1 \le j < k \le 2n \\ j+k \ne 2n+1}} |A_j A_k|}.$$
Since $A_1A_{n+1}, \dots, A_nA_{2n}$ are the longest diagonals (chords) in the regular $2n$-gon $A_1 \cdots A_{2n}$,
$$\prod_{\substack{1 \le j < k \le 2n \\ k-j \ne n}} |A_j A_k| = \left( \prod_{j=1}^{n-1} 2 \sin \frac{j\pi}{2n} \right)^{2n} = (\sqrt{n})^{2n} \ge 1,$$
which implies
$$\prod_{1 \le j < k \le n} (x_j - x_k) \ge \left(\frac{1}{2}\right)^{\frac{n(n-1)}{2}}.$$
In conclusion, $C = \frac{1}{2}$. $\square$