I'm wondering if every closed surface can be realized as a mean convex hypersurface in $\mathbb{R}^3$, i.e. the mean curvature vanishes or points inward.
Categorizing by genus: for $S^2$ ($g = 0$) this is evidently true, and for $T^2$ ($g = 1$), one can take the radius of the tube to be very small relative to the outer radius (see e.g. here) to force mean convexity. I have no idea if this is true for higher genus.
Yes, every closed orientable surface can be embedded in $R^3$ with positive mean curvature.
One way to construct higher genus examples is by gluing thin tori, which are mean convex. For instance we can take a pair of these tori and position them next to each other, like a pair of doughnuts on a table top. Then one connects them by constructing a neck as follows.
Let $o$ be the midpoint of the line segment which connects the closest points of the tori. Consider the double cone with vertex at $o$ which grazes each of the tori, i.e., the cone consisting of all lines passing through $o$ which are tangent to the tori. By gluing a portion of the cone to portions of the tori we obtain a compact surface with a singularity at $o$ which is otherwise $C^1$.
The singularity at $o$ can be smoothed with positive mean curvature provided that the tori are sufficiently far apart. Furthermore we may smoothen the surface along the curves where the cone meets the tori, again with positive mean curvature, by slightly inflating the cone. This is possible because the curves where the cone and the tori meet lie on the portions of the tori with positive Gauss curvature (the non-positively curved parts of the tori are not visible from $o$).
