Note: This is an optimalization problem. You can scroll to below to see what needs to be optimalized
I am trying to prove the isoperimetric theorem for dodecahedra, hehahedra and dodecahedra, the polyhedra where the points meet in only 3 faces. These are known to be the easiest to prove (the case for isocahedra hasn't even been proven at all, from my understanding). The theorem reads as follows:
Among all [x]hedra with the same surface area, the regular [x]hedron has maximal volume.
To start off, we have the Lindemann-Minkowski theorem, which reads as follows:
Let $S$ be any $n$-dimensional polyhedron, and let $S_L$ be the polyhedron that has an inscribed sphere of unit radius, and that has the ssurface normals as $S$. The functions $V$ and $A$ denote the volume and surface area respectively. Then we have that:\begin{align*}\frac{A(S_L)^n}{V(S_L)^{n-1}} \leqslant \frac{A(S)^n}{V(S)^{n-1}}\end{align*} With equality if and only if $S$ and $S_L$ are homothetic
This means that we can limit ourselves to polyhedra with an inscribed sphere of radius $r$. For such polyhedra, we have $V(S)=r\frac{1}{3}A(S)$. As such, we can rewrite the theorem as follows:
Among all [x]hedra with an inscribed sphere with radius $1$, the regular [x]hedron has a minimal volume.
In order to prove this, let $S$ be a tetra-/hexa-/dodecahedron where the inscribed sphere is of unit length. Let $O$ be the center the sphere. Pick a vertex $T$ of $S$. Let $A$, $B$ and $C$ be the tangent points of the adjacent faces. Let $C(O, ABC)$ be the cone through $O, A, B, C$ (where $O$ is the top). Let $X, Y, Z$ be the points of $\delta C(O, ABC)$ that lie on two faces. We get the following:
We now compute the volume of the convex hull $F_{i}=H(A, Z, T)$ (where $i$ is the tuple $(T,A,Z)$). Let $\alpha_i = |A\hat{O}Z|$ and $\beta_i = |T\hat{O}Z|$. Then we get:
\begin{align*} V(F_i) = \frac{1}{2}|AZ|*|TZ|=\tan(\alpha_i)1*\tan(\beta_i)|ZO|=\tan(\alpha_i)\tan(\beta_i)\sec(\alpha_i)\end{align*}
Our job is to minimize $\sum_i V(F_i)$ where the $i$ are all possible tuples. (there are $6$ of those in $C(O,A,B,C)$). The solid angle of $H(A,Z,T,O)$ at $O$ is given by $\frac{4\pi}{(2\pi^2)}\alpha_i\beta_i$. As such, $\sum_i\alpha_i\beta_i$ takes a given value.
Now let $\gamma_i = |Z\hat{A}T|$. The $\gamma_i$ on the same face sum up to $2\pi$. Since the amount of edges per face and the amount of faces is fixed, it follows that $\sum_i\gamma_i$ is constant as well. But we have that: \begin{align*}\gamma_i = \arctan(\frac{|TZ|}{|AZ|})=\frac{\sec\alpha_i\tan\beta_i}{\tan\alpha_i}=\frac{\tan\beta_i}{\sin\alpha_i}\end{align*}
So the problem becomes: Minimalize:
\begin{align*}A(S)=\frac{1}{2}\sum_i{\tan(\alpha_i)\tan(\beta_i)\sec(\alpha_i)}\end{align*} Subject to the conditions: \begin{align*}\sum_i\alpha_i\beta_i = C_1\end{align*} \begin{align*}\sum_i\arctan(\frac{\tan\beta_i}{\sin\alpha_i}) = C_2\end{align*}
The question is: How should I proceed here? And is this even the correct way of tacking this problem at all?
EDIT: I am aware that $\tan x$ is a convex function, i.e. $\tan(tx + (1-t)y) \leqslant t\tan(x) + (1-t)\tan(y)$ for $t\in (0,1)$, with equality only if $x=y$. For polygons, the problem becomes very easy to solve with this, but I just cannot find a way to apply it here.