If a simply connected set $\Omega\subset\mathbb{R}^n$ has $C^2$ boundary such that the mean curvature $H$ of $\partial \Omega$ satisfies:
$$H\geq 1$$
Does this imply that $\Omega\subset B_1$ after rigid transformation? I believe this is true but did not find it by searching.
Edit: As pointed out by @anything in the comment below, the claim may be false in $\mathbb{R}^n$ for $n\geq 3$. But what if $n=2$? Would this be true?
The result is true in two dimensions. Instead of using the mean curvature, we will use the equivalent geodesic curvature of the boundary. The proof goes via the Gauss map.
Step 1: basic reductions
Let $\Omega$ be a bounded open set with $C^2$ boundary in $\mathbb{R}^2$. By compactness (ask if you need more details) there exists an open ball $B\supseteq \Omega$ of minimum radius, in the sense that any $B'$ with strictly smaller radius cannot contain $\Omega$.
We use the following basic geometric fact:
there exists at least two points on $\partial\Omega$ where it is tangent to $\partial B$.
Sketch of proof: If there is only one point, then shrinking $B$ slightly toward the tangent point will result in a $B'$ that is smaller and still contains $\Omega$.
Since $\Omega$ is assumed to be simply connected, we have that $\partial\Omega$ is a closed curve, which we call $\gamma$ for convenience. Denote by $T$ the unit tangent and $N$ the unit inward normal along $\gamma$. The geodesic curvature is $\langle N, \nabla_TT \rangle =: \kappa$. That $\kappa$ is assumed to be signed implies that the map $T:\gamma\to\mathbb{S}^1$ is a $C^1$ diffeo, and so is invertible. This gives us (essentially) the Gauss map.
We can do the same thing with the curve $\partial B = \mu$.
The discussion above means that there exists a parametrization such that we can write $\gamma,\mu: \mathbb{S}^1\to\mathbb{R}^2$ such that the unit tangent vector fields $$ \frac{\dot{\gamma}}{|\dot{\gamma}|} = \frac{\dot{\mu}}{|\dot{\mu}|} = (-\sin(\theta),\cos(\theta)) $$ (so that the unit inward normals are $(-\cos(\theta),-\sin(\theta))$.
Step 2: relating curvature and speed
The main observation is that this parametrization is chosen so that the geodesic curvature is the inverse of the curve speed. We see this for $\gamma$: the derivative $$ \nabla_T T = \frac{1}{|\dot{\gamma}(\theta)|} \partial_\theta \frac{\dot{\gamma}(\theta)}{|\dot{\gamma}(\theta)|} = \frac{1}{|\dot{\gamma}(\theta)|} N$$ and so we see that $\kappa |\dot{\gamma}| = 1$.
Step 3: revisiting the tangent points
As argued before, $\partial\Omega$ and $\partial B$ are tangent at at least two points. Note that at these two points both the location and unit tangent of the boundary curves are equal. Using the Gauss map, this means that we can find two points $\theta_0, \theta_1$ such that $\gamma(\theta_0) = \mu(\theta_0)$ and $\gamma(\theta_1) = \mu(\theta_1)$. Without loss of generality we can assume that $\theta_0 = 0$ and $\theta_1 \in (0,\pi]$.
Therefore we have $$ \int_0^{\theta_1} \dot{\gamma}(\theta) ~d\theta = \int_0^{\theta_1} \dot{\mu}(\theta) ~d\theta $$ Let $\rho$ denote the radius of $B$, then using that $\dot{\gamma} = |\dot{\gamma}| (-\sin(\theta),\cos(\theta))$ (and similarly for $\mu$) we have $$ \int_0^{\theta_1} \frac{1}{\kappa(\theta)} \sin(\theta) ~d\theta = \rho \int_0^{\theta_1} \sin(\theta)~d\theta. $$ Using that $\sin(\theta)$ is signed on $(0,\pi)$, we see that $\rho$ is a (weighted) average of $\frac{1}{\kappa}$. As we assumed that $\kappa \geq 1$ everywhere, we have $\rho \leq 1$, proving our claim.
A comment on the proof:
The argument above uses that $\kappa \geq 1$ on a "large" subset of $\gamma$ (through integration); so that it uses some "non-local information". This is the key part of the argument.
Using only local information you get a weaker result. Choose $x_0\in \Omega$ and let $y_0$ be a point on $\partial\Omega$ that minimizes the distance to $x_0$. Then locally taking polar coordinates centered at $x_0$ we find that the portion of $\partial\Omega$ near $y_0$ may be parametrized as $$ (r(\theta),\theta) $$ with $\theta\in (a,b)$ and $\theta_0$ corresponding to $y_0$. Since $y_0$ is a point of closest approach we have $r'(\theta_0) = 0$. A computation then shows that the geodesic curvature at $y_0$ must be $$ r(\theta_0) \kappa(y_0) = 1 - \frac{r''(\theta_0)}{r(\theta_0)} $$ As $\theta_0$ is a local minimum we have $r'' \geq 0$ and hence $$ r(\theta_0) \kappa(y_0) \leq 1 $$ and $$ r(\theta_0) \leq \frac{1}{\kappa(y_0)} $$
So this "pointwise maximal principle" argument only gives the weaker result
If $x\in \Omega$, then $\mathrm{dist}(x,\partial\Omega) \leq 1$.
