Let $X$ be a bounded subset of $\mathbb R^3$, and denote by $Conv(X)$ the convex hull of $X$, i.e, the smallest convex subset of $\mathbb R^3$ containing $X$. As $X \subset Conv(X)$ by definiton, the diameter-inequality $diam(X) \leq diam(Conv(X))$ certainly holds, and I wonder if a partial converse is also true:
There exists an integer $n \in \mathbb N$ and a constant $c > 0$, such that the following holds: If $\Gamma \subset \mathbb R^3$ is a (smooth) simple closed curve, we have the inequality \begin{equation} diam(Conv(\Gamma)) \leq c*diam(\Gamma)^n \end{equation}
It can be shown that any (smooth) simple closed curve $\Gamma$ in $\mathbb R^3$ bounds a disc $D$ with $Area(D) \leq (4\pi)^{-1}*length(\Gamma)^2$. However, things get a bit more complicated when passing to the diameter - it is possible to construct a sequence $(D_k)_{k=1}^\infty$ of discs bounded by $\Gamma$ whose diameter grows arbitrarily large as $k \to \infty$, but have uniformly bounded Area. Moreover, the inequality $diam(\Gamma) \leq lenght(\Gamma)$ can easily shown to be true, but it is not possible to find simliar inverse inequality (as can be seen by considering curves inside a small ball that are of large variation). Thus, it seems highly unlikely that the Area-length inequality can be used to relate the diameter of a disc with the diameter of a corresponding boundary curve.
However, when considering a least area disc $D$, bounded by $\Gamma$, we have a particularly well-behaved disc, such that $D \subset Conv(\Gamma)$. Therefore, the inequality in question could be used to establish an analogous relation between the diameter of such a disc with its boundary curve.
Taking the convex hull does not increase the diameter:
$$\textrm{diam}\ ( Conv(X)) = \textrm{diam} (X)$$
This follows from the fact that $$d( P, \lambda Q_1 + (1-\lambda) Q_2) \le \lambda d (P, Q_1) + (1-\lambda) d (P, Q_2)$$ as the distance is given by a norm.