how prove this inequality $\sum\limits_{1\le i<j\le n}|z_{i}-z_{j}|\le n\cot{\frac{\pi}{2n}}$

Let complex $z_{i}(i=1,2,\cdots,n$),and such $|z_{i}|=1$

show that $$\sum_{1\le i<j\le n}|z_{i}-z_{j}|\le n\cot{\dfrac{\pi}{2n}}$$

I tried C-S and complex numbers Lagrange's identity, but without success.

Label the points $A_1, \ldots, A_n, A_{n+1} = A_1, \ldots, A_{2n-1} = A_{n-1}$ in order. Let $A_{i,j}$ denote the measure of the arc from $A_i$ to $A_j.$ Then the problem reduces to proving $$2\sum_{i=1}^{n} \sum_{j=1}^{n-1} \sin \frac{A_{i,i+j}}{2} \le n\cot\frac{\pi}{2n}.$$

Note each of the angles in question is less than $\pi,$ so we can apply Jensen's inequality for fixed $j$ to get $$\frac{1}{n}\sum_{i=1}^{n} \sin\frac{A_{i,i+j}}{2} \le \sin\left(\sum_{i=1}^{n} \frac{A_{i,i+j}}{2n}\right) = \sin \frac{\pi j}{n}.$$ Now do the sum over $j$ to finish.