The well-known isoperimetric inequality in $\mathbb R^2$ states that for a $\Gamma \subset \mathbb R^2$ a curve (simple, closed, piecewise $C^1$) and $A$ the area of the bounded component of $\Gamma^c$, then $4\pi A \leq l^2$, where $l$ denotes the lenght of $\Gamma$.
Now adding one dimension to the ambient space, a simple closed curve $\Gamma \subset \mathbb R^3$ bounds an infinite number of compact surfaces and it is easy to see that there is no upper bound on the area of such surfaces, as any such bound can be exceeded by pushing the inside of a disk with boundary $\Gamma$ sufficiently far away from $\Gamma$. However, there seems to be a well-known, albeit much weaker version of this inequality in $\mathbb R^3$, which states:
Let $B \subset \mathbb R^3$ be a bounded convex domain. Then, there exists a constant $C_B > 0$ such that for any simple, closed $C^1$ curve $\Gamma \subset B$, there is a disk $D \subset B$ with $A(D) \leq C_B L(\Gamma)^2$
Having read that this is a particularily easy statement to prove, I have tried looking for the class of disks with a fixed boundary curve which are easiest to parametrize. In this regard, there seem to be no simpler class than cones, i.e, disks of the form $g(t,s) := tp + (1-t)f(s)$, where, $t \in [0,1]$ $p \in B$ is a fixed point and $f:[0,1] \to \mathbb R^3$ is a parametrization of $\Gamma$. By convexity of $B$, the image of $g$ completely lies in $B$. Also, since $B$ is bounded, we have $d:= diam(B) < \infty$. Then, we can compute \begin{equation} A(D) = \int_{[0,1]^2} ||\frac{\partial g}{\partial t} \times \frac{\partial g}{\partial s}|| dtds = \int_{[0,1]^2} \sqrt{(||\frac{\partial g}{\partial t}||\frac{\partial g}{\partial s}||)^2 - \langle\frac{\partial g}{\partial t},\frac{\partial g}{\partial s}\rangle^2}dtds \leq \int_{[0,1]^2} ||\frac{\partial g}{\partial t}||||\frac{\partial g}{\partial s}||dtds = \int_{[0,1]^2}(1-t)||p-f(s)||||f'(s)||dtds \leq d\int_{[0,1]}||f'(s)||ds = dL(\Gamma)\end{equation}
Since the expression $L(\Gamma)$ on the right-hand side is missing a square, this is not the inequality I was originally looking for. Intuitively, it would make much more sense for $L(\Gamma)$ to be squared, but I do not see how I can achieve this by a simple calculation. Can anyone help me out here ?
I just came up with the solution, the trick is just to choose the apex $p$ of the cone to be sufficiently close to $\Gamma$ in order to get that $\sup_{q \in \Gamma} ||p-q|| \leq L(\Gamma)$. Going through the same reasoning as in the original calculations, one comes to the following conclusion:
For $n \geq 2$ and any simple, closed, piecewise $C^1$ curve $\Gamma \subset \mathbb R^n$, there is a disc $D \subset \mathbb R^n$ with $\partial(D) = \Gamma$ and $A(D) \leq \frac{1}{2}L(\Gamma)^2$
This is, by no means, the optimal inequality, as it is well-known that there exist discs with boundary $\Gamma$ and $A(D) \leq \frac{1}{4\pi} L(\Gamma)^2$ (which is optimal, as can be seen by comparing the area and the perimeter of the unit disc in $\mathbb R^2$). However, the latter result relies on much deeper analysis, such as the existence of minimal surfaces under fixed boundary conditions and is, in return, much harder to prove.