TITLE: Prove that the sphere is the only closed surface in $\mathbb{R}^3$ that minimizes the surface area to volume ratio.
QUESTION [16 upvotes]: It is well known that a sphere minimizes the surface area to volume ratio since it reaches equality in the Isoperimetric Inequality. I'm trying to prove that no other closed surface has this property.
Since the inequality says that $36\pi V^2 \leq A^3$ we can state the problem as:
Let $\mathcal{S}$ be a simple closed surface in $\mathbb{R}^3$ with surface area $4\pi$. Prove that if its volume is $\dfrac{4}{3}\pi$ then $\mathcal{S} = \mathbf{S}^2$.
My hypothesis:
If there is a surface $\mathcal{S}\neq \mathbf{S}^2$ but with the same surface area to volume ratio then there is a function $F(t,x):I\times \mathbf{S}^2\to \mathbb{R}^3$ so that:
$$F(t,x_0) \text{ is continuous }\forall x_0\in\mathbf{S}^2$$
$$F(t_0,\mathbf{S}^2) \text{ is a simple closed surface }\forall t_0\in I$$
$$F(0,\mathbf{S}^2) = \mathbf{S}^2\qquad F(1,\mathbf{S}^2) = \mathcal{S}$$
$$F(0,\mathbf{S}^2) \cong F(t,\mathbf{S}^2)\qquad \forall t\in I$$
$$V(F(t,\mathbf{S}^2)) = \frac{4}{3}\pi\qquad A(F(t,\mathbf{S}^2)) = 4\pi\qquad \forall t\in I$$
Where $\cong$ refers to equivalence via a bijective function, and $V(X)$ and $A(X)$ are, respectively, the volume and surface area functions.
In other words, there is some kind of continuous homeomorphism between both surfaces that conserves both volume and surface area.
Then by using variation calculus we can prove that if a surface is "almost" a sphere, it has a higher $S/V$, thus there cannot be such function $F$ and therefore there is no such surface $\mathcal{S}$.
Perhaps my hypothesis is false. Feel free to prove the statement without using it.

REPLY [5 votes]: The problem of to show that the surface with minimal area that enclose a given volume is a sphere is not at all simple.
Using the calculus of variation we can show that the surface must have constant curvature. I sketch the proof:
Given a smooth closed surface $S$ in $\mathbb{R}^3$, let $A$ its area and $V$ the volume enclosed.
As varied surface consider the surface $S(t)$ obtained by displacing each point of $S$ by a vector $th\vec n$, where $\vec n$ is the normal to the surface and $h$ is  a smooth real valued function $h:S\rightarrow \mathbb{R}$. The first variations of the area and the volume are:
$$
A'(0)=-\int_ShHdA \qquad V'(0)=\int_ShdA
$$
where $H$ is the mean curvature of the surface with respect to the normal.
Now, to have the minimum area for given volume, it must be true that whenever $V'=0$ also $A'=0$, so $H$ must be a constant on $S$.
But now the question is to show that a surface of constant mean curvature is necessarily a sphere and this can be proved, but the proof is not simple.
A first proof was given by Liebmann in 1900 for compact strictly convex surfaces, generalized by Aleksandrov in 1962 to surfaces with some kind of self-intersection. But, as far as i know, the question of the possible existence of surfaces of any topological type with self intersection and constant mean curvature, is unanswered.
For a complete treatment of this topic you can see here.