Let $S$ be the surface of a convex body, polyhedral or smooth, 
embedded in $\mathbb{R}^3$.
For a point $x \in S$, let $F(x)$ be the set of furthest points
from $x$, measured by shortest paths on the surface $S$.
Let $f(x)$ be the length of those shortest paths:
$|x y|$ for $y \in F(x)$.


It seems natural to hope that



> 
> **Hypothesis**: For any $x \in S$, $f(x) \ge \tfrac{1}{2} \mathrm{diam}(S)$.
> 
> 
> 


Here $\mathrm{diam}(S)$ is the maximum distance between any two points
on $S$ (again measured by shortest paths on the surface of $S$).
Suppose, for example, that $\rho$ is a diameter-realizing geodesic.
Then for any $x \in \rho$, $f(x) \ge \tfrac{1}{2} |\rho|$, just
tracking along $\rho$. 


A non-comprehensive literature search has
failed to uncover a relationship between $f(x)$ and $\mathrm{diam}(S)$.




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Itoh, Jin‐ichi and Costin Vǐlcu. "Criteria for farthest points on convex surfaces." *Mathematische Nachrichten* 282, no. 11 (2009): 1537-1547.
[Journal link](https://onlinelibrary.wiley.com/doi/abs/10.1002/mana.200610811).



Furthest distance half the diameter?
Denote the diameter by $d$ and distance by $|x-y|$. Then there
are $y,z$ such that $d=|y-z|$ and we have by triangle inequality for every $x$:
$$d=|y-z|\leq |y-x|+|x-z|\leq 2f(x),$$ 
so we obtain your inequality. Notice that I did not use convexity, or any other of your assumptions, only the
triangle inequality.


