weaker than isoperimetric inequality for surfaces in $\mathbb{R}^3$

Question

I'm trying to prove, without the aid of the isoperimetric inequality, that for any compact manifold with given surface area $S$ in $\mathbb{R}^3$, that the volume $V$ must be bounded. By this I mean that any sequence of homeomorphisms that also preserve $S$, can't result in an unbounded volume $V$.

In particular, I'm searching for an elementary proof of a weaker-than-isoperimetric inequality:

$$ V \leq f(S) \tag{*}$$

I tried to find such an inequality using the Pappus Centroid Theorem but I have not been successful so far.

Note 1: I am interested in a result that generalises easily to compact manifolds in $\mathbb{R}^n$ that aren't necessarily differentiable.

Note 2: My motivation comes from a problem in mathematical physics where an isoperimetric inequality is deduced so I don't want to assume the isoperimetric inequality beforehand.

Answer 1

After discussing this with Tony Carbery, an analysis professor at Edinburgh, I realised that the Loomis-Whitney inequality answers this problem adequately. In their original paper which is only two pages long they use cubical projections of n-space to demonstrate the following theorem:

Let $m$ be the measure of an open subset O of $\mathbb{R}^n$, and let $m_1,..m_n$ be the (n-1)-dimensional measures of the projections of O on the coordinate hyper-planes. Then:

$$m^{n-1} \leq \prod_{i=1}^{n} m_i \tag{*}$$

From this we may deduce that if $M$ is a compact manifold in $\mathbb{R}^3$ with given surface area $S$:

$$V \leq S^{3/2} \tag{**}$$

This is precisely the result I wanted.

Note 1: I can reproduce the details of their proof here but I think that the original paper is worth reading.

Note 2: This is clearly weaker than the sharp isoperimetric bound $V \leq \frac{S^{3/2}}{3 \sqrt{4 \pi}} \huge$ which is much more difficult to prove.