Proving that out of all the possible n-gons that exist in the unit circle, the one with the maximum possible perimeter is the regular $n$-gon. So in the context of my Convex Analysis studies, I have come across this problem:

*

*First I have to prove that $ - \sin x $ is convex over $[0, \pi]$. That's easy enough using the second derivative theorem.

*Now, using the answer from above, I have to prove that  out of all possible $n$-gons that can be specified on the unitary circle, the one with the maximum possible perimeter is the normal $n$-gon.

My geometric intuition is not good, and I cannot figure out how to do this using the tools from Convex Analysis. Any help would be greatly appreciated.
 A: I can't really improve on @person's answer other than some convex stuff.
The problem is $\min \{ N(x) | 0 \le x_k \le \pi, \sum_k x_k = 2 \pi \}$, where
$N(x) = -\sum_k 2 \sin { x_k \over 2} $. Given the restriction on the arguments, we see that $N$ is convex. Since the feasible set is compact, there is a solution. Note that if $x$ is a solution, then so is $\Pi x$ for any permutation $\Pi$. Let ${\cal P}$ be the set of permutations. Since $N$ is convex, we have
$N( {1 \over |{\cal P}|}\sum_\Pi \Pi x) \le {1 \over |{\cal P}|} \sum_\Pi N(\Pi x) = N(x)$, and so a solution of the form $(\theta,...,\theta)$ exists ($\theta = {2 \pi \over n}$, of course).
A little more work shows that $N''>0$ and so $N$ is strictly convex, hence the solution is unique.
A: Convexity Approach
Let $\theta_k$ be the angle subtended by side $k$ as viewed from the center of the circle. The perimeter of the polygon is
$$
2\sum_{k=1}^n\sin(\theta_k/2)
$$
$\sin(x)$ is concave on $[0,\pi]$. This means that
$$
\sin\left(\frac{\theta_k/2+\theta_{k+1}/2}2\right)\ge\frac{\sin(\theta_k/2)+\sin(\theta_{k+1}/2)}2
$$
If two adjacent sides are not equal, then replacing them with equal sides subtending the average of their angles will not decrease the perimeter. Thus, we see that a regular polygon will give the maximum perimeter.

Variational Approach
For simplicity of notation, let the indices be cyclic; i.e. $x_0=x_n$ and $x_1=x_{n+1}$.
For all $\delta x_k$ that maintain $|x_k|=1$, that is,
$$
x_k\cdot\delta x_k=0\tag1
$$
we want
$$
\begin{align}
0
&=\delta\sum_{k=1}^n|x_k-x_{k-1}|\tag{2a}\\
&=\sum_{k=1}^n\frac{x_k-x_{k-1}}{|x_k-x_{k-1}|}\cdot\delta(x_k-x_{k-1})\tag{2b}\\
&=\sum_{k=1}^n\frac{x_k-x_{k-1}}{|x_k-x_{k-1}|}\cdot\delta x_k-\sum_{k=0}^{n-1}\frac{x_{k+1}-x_k}{|x_{k+1}-x_k|}\cdot \delta x_k\tag{2c}\\
&=\sum_{k=1}^n\left(\frac{x_k-x_{k-1}}{|x_k-x_{k-1}|}+\frac{x_k-x_{k+1}}{|x_{k+1}-x_k|}\right)\cdot\delta x_k\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a)}$: to maximize the perimeter, the variation of the perimeter must be $0$
$\text{(2b)}$: chain rule
$\text{(2c)}$: split the sum and reindex the second sum
$\text{(2d)}$: apply the cyclic indices and combine the sums
Orthogonality says that to get $(2)$ for all $\delta x_k$ that satisfy $(1)$, there must be a constant $\lambda$ so that
$$
\begin{align}
\lambda x_k
&=\frac{x_k-x_{k-1}}{|x_k-x_{k-1}|}+\frac{x_k-x_{k+1}}{|x_{k+1}-x_k|}\tag{3a}\\[3pt]
&=2\cos(\theta_k/2)\,x_k\tag{3b}
\end{align}
$$
$\text{(3a)}$: compare $(1)$ and $(2)$
$\text{(3b)}$: the sum of two unit vectors is a vector of length
$\phantom{\text{(3b):}}$ $2\cos(\theta/2)$ in the direction of their bisector,
$\phantom{\text{(3b):}}$ where $\theta$ is the angle between the unit vectors

Since $2\cos(\theta_k/2)=\lambda$, we know that $\theta_k$ is the same for all $k$. This makes each triangle $(0,x_k,x_{k+1})$ isosceles and congruent. Thus, the polygon is regular.
A: Unfortunately I do not know Convex Analysis too well, but here is one possible answer using Jensen's inequality. Each one of the sides of a polygon must be a chord of the circle, and each chord has an angle which corresponds to it.

Consider $n$ such angles, $\{\theta_1,\theta_2, \dots ,\theta_n\}; \sum_{k=1}^n \theta_k = 2\pi; \theta_i < \pi$ , which correspond with these chords, which form the perimeter of a polygon. It is not too hard to prove that the length of each chord is $2\sin(\theta_i/2)$.
The perimeter would be the sum of the chords:
$$2\sum_{k=1}^{n}\sin(\theta_k/2)=P$$
Since sine is concave from 0 to $\pi$, we may use the concave version of Jensen's:
$$f(\frac1n\sum_{k=1}^nx_i)\geq\frac1n\sum_{k=1}^nf(x_i)$$
$f(x)=2\sin(x/2), x_i=\theta_i$:
$$2\sin(\frac1{2n}\sum_{k=1}^n\theta_k))\geq\frac1n2\sum_{k=1}^{n}\sin(\theta_k/2)$$
$$P=2\sum_{k=1}^{n}\sin(\theta_k/2)\leq2n\sin(\frac1{2n}\sum_{k=1}^n\theta_k))$$
$\sum_{k=1}^n \theta_k = 2\pi$:
$$P=2\sum_{k=1}^{n}\sin(\theta_k/2)\leq2n\sin(\pi/n)$$
Now the current RHS is exactly what would happen if all of the $\theta_i$s were equal, and since every other case is lesser than that, we can see that the maximum possible perimeter is the regular polygon.
EDIT:
To prove that the length of the cord is $2 \sin(x/2)$ consider the following figure:

Where, given a circle of radius 1, we want to find the cord $CD$.
$AD=\cos(x)$, and so $DB=1-\cos(x)$.
$ABC$ is isosceles, so $\angle ACB=\angle ABC=\frac{\pi-x}2$
$\frac{DB}{\cos(\angle ABC)}=CB$, so substituting everything in, and using the double angle identity:
$$CB=\frac{1-\cos(x)}{\cos(\frac{\pi-x}2)}=\frac{1-(1-2\sin^2(x/2))}{\sin(x/2)}=\frac{2\sin^2(x/2)}{\sin(x/2)}=2\sin(x/2)$$
