If we have a set of points in the plane of diameter 1, the perimeter of its convex hull is at most $\pi$; this is attained by the shapes of constant width (circle, Releaux triangle, etc). See Barbier's theorem.
(I can find this fact mentioned without citation in Gallego and Solanes, Perimeter, diameter, and area of convex sets in the hyperbolic plane, but have not located a proof; direction to such would be appreciated. This math.SE post asks the same question, though I'm not sure the accepted answer's argument is fully rigorous.)
I'm curious if this result extends to higher dimensions - for instance, among convex solids of diameter $1$ in $\mathbb{R}^3$, is the surface area of the solid at most $\pi$?
Interestingly, it seems that in $\mathbb{R}^3$ the solids of constant width do not all attain the same value: this source suggests that Meissner bodies of unit diameter have a surface area of only $2.934$. It also notes that among solids of a fixed constant width, volume and surface area are monotonically increasing in each other, so the constant-width surface of maxmimal volume will also be that of maximal surface area. (I believe this is the sphere, but I don't know if this has been proven beyond the case of solids of revolution of constant width.)
This disparity suggests to me that proof techniques used in the two-dimensional case may not generalize, and that there is some chance that non-spherical shapes (perhaps other solids of constant width?) have higher surface area. The isoperimetric inequality implies that any shape of unit diameter with higher volume than a ball would serve as such a counterexample, but I believe no such shapes exist.