How to give a tight upper bound for the spectral norm of a Cauchy matrix?

Recently, I came across a problem where I needed to give an upper bound of the spectral norm of a Cauchy matrix $$C$$.

$$C$$ is a $$2n \times 2n$$ matrix with element $$c_{ij}$$ in the form $$$$c_{ij} = \frac{1}{x_i-y_j}, \quad 1 \leq i \leq 2n, 1\leq j \leq 2n,$$$$ where $$x_1,x_2,\cdots,x_{2n}$$ are $$2n$$ distinct real numbers in the interval $$[-1,1]$$, and $$y_1,y_2,\cdots,y_{2n}$$ are $$2n$$ distinct roots of equation $$z^{2n}+1 = 0$$.

My problem is that is there a real number $$h(n)$$(for example, $$2n^{3/2}$$) as small as possible that depends only on $$n$$ such that $$$$\|C\| = \sigma_{\max}(C) = \sup_{v \in \mathbb{C}^{2n},\|v\|=1} \|Cv\| \leq h(n)?$$$$

I have obtained that $$$$\|C\| \leq \sqrt{2n}\|C\|_1 = \sqrt{2n} \max_{j=1,2,\cdots,2n} \sum_{i=1}^{2n}\left|c_{ij} \right| = \sqrt{2n} \max_{j=1,2,\cdots,2n} \sum_{i=1}^{2n} \left|\frac{1}{x_i-y_j} \right| \leq \sqrt{2n} \max_{j=1,2,\cdots,2n} \sum_{i=1}^{2n} \left|\sin \frac{\pi}{2n} \right|^{-1} \approx \frac{4\sqrt{2}n^{5/2}}{\pi},$$$$ but this upper bound is too loose.

The Cauchy-Schwartz inequality gives the following estimate for the spectral norm $$\lVert C \rVert \leq \left(\sum_{i = 1}^n \sum_{j = 1}^n \lvert C_{ij} \rvert^2\right)^{1/2}.$$ Now plug in the bound $$\lvert{C_{ij}\rvert} \leq\frac{1}{\lvert{\Im y_j\rvert}}$$ and the desired estimate should follow easily.