I'm trying to prove the following statement
Prove that the diameter is lower or equal to the half of the perimeter in a closed, convex and planar curve.
Is my proof correct?
$\textbf{My attempt:}$
Let $\gamma: I \longrightarrow \mathbb{R}^2$ a closed and convex planar curve and $D$ the diameter of $\gamma(I)$ considering the Euclidean distance $d$.
Since $\gamma(I) \subset \mathbb{R}^2$ is compact, we know that there is a closed ball with radius $r$ containing $\gamma(I)$ and such that $r \leq \frac{D}{\sqrt{3}}$ by Jung's theorem in the plane
Clearly, $\frac{D}{4} \leq \frac{D}{\sqrt{3}}$, then we choose $r = \frac{D}{4}$ and we have that the diameter of $\gamma(I)$ is lower or equal to diameter of a closed ball, i.e.,
$$D \leq 2r = 2 \frac{D}{4} = \frac{D}{2},$$
but the diameter of $\gamma(I)$ is lower or equal to the perimeter $L$ of $\gamma(I)$, i.e., $D \leq L$, then
$$D \leq 2r = 2 \frac{D}{4} = \frac{D}{2} \leq \frac{L}{2},$$
which concludes that $D \leq \frac{L}{2} \ \square$.
Thanks in advance!
I didn't understand your attempt; in particular when you say "then we choose $r={D\over4}$". In my view we don't need Jung's theorem.
Since $\gamma(I)$ is compact there are two points $p$, $q\in\gamma(I)$ with $|p-q|=D$. These two points divide $\gamma$ into two parts $\gamma'$, $\gamma''$, each of them connecting $p$ with $q$, hence of length $\geq |p-q|=D$. It follows that $L(\gamma)=L(\gamma')+L(\gamma'')\geq 2D$.