Area of a cyclic polygon maximum when it is a regular polygon

My question: Let $n$ points $A_1, A_2,\ldots,A_n$ lie on given circle then show that $\operatorname{Area}(A_1A_2\cdots A_n)$ maximum when $A_1A_2\cdots A_n$ is an $n$-regular polygon.

Let's call the center of the circle $O$.

First, it's clear that none of the angles $\angle A_1OA_2$, $\angle A_2OA_3$,…,$\angle A_nOA_1$ should be greater than $\pi$. If they were, the polygon would be smaller then half of circle $O$. Obviously, this is not maximal.

Let's call $m\angle A_1OA_2=\theta_1$, $m\angle A_2OA_3=\theta_2$,…, $m\angle A_1OA_n=\theta_n$.

The area of the polygon is: $$\frac{r^2}{2}\sum_{i=1}^n \sin\left(\theta_i\right)$$ Note that $\sin(x)$ is concave on $[0,\pi]$. From Jensen's inequality: $$\sin\left(\frac{1}{n}\sum_{i=1}^n \theta_i\right)\geq \frac{1}{n}\sum_{i=1}^n \sin\left(\theta_i\right)$$ We know that $\displaystyle \sum_{i=1}^n \theta_i=2\pi$ so we can deduce the following: $$n\sin\left(\frac{2\pi}{n}\right)\geq \sum_{i=1}^n \sin\left(\theta_i\right)$$

This means that: $$\frac{r^2}{2}\sum_{i=1}^n \sin\left(\theta_i\right)\leq \frac{nr^2}{2}\sin\left(\frac{2\pi}{n}\right)$$

This suggests that the area of the polygon is maximized when $\theta_1, \theta_2, \ldots, \theta_n = \displaystyle\frac{2\pi}{n}$.

My answer: Show that if the polygon is not regular, then you can get a larger area by moving a sigle vertex along the circle.

This answer is different from my previous answer, and maybe even better.

If our polygon is not regular, we can find $A_{i-1}$, $A_i$, and $A_{i+1}$, such that $A_i$ is not the midpoint of the arc between $A_{i-1}$ and $A_{i+1}$. Notice that when we move $A_i$ to the midpoint of this arc, the area of the polygon becomes bigger, since the area of $\triangle A_{i-1} A_i A_{i+1}$ increases. Ergo, the maximal polygon is the polygon for which $A_i$ is the midpoint of arc $A_{i-1}A_{i+1}$ for all $i$.