I'm try to proof that the area of a convex closed regular simple plane curve is lower or equal to width times the diameter ($A \leq w D$). Intuitively, it's clear that $A \leq w D$, because the convex closed regular simple plane curve can be enclosed by a rectangle with base length $D$ and height length $w$ as shown in the figure below
I would like to know how can I proof that $A \leq w D$. I tried to proof by Isoperimetric Inequality, but I just prove with this that $A \leq L D$ as shown in my attempt below.
$\textbf{My attempt:}$
The curve is a Jordan's Curve, then we can use the Isoperimetric Inequality:
$$4 \pi A \leq L^2,$$
where $A$ is the area enclosed by the curve and $L$ is the length of the curve
Because the diameter is the maximum of width of a curve ($L \leq D$), we have that
$$L^2 \leq LD,$$
then
$$A \leq \frac{L^2}{4 \pi} \leq \frac{LD}{4 \pi} \leq LD$$
Thanks in advance!
$\textbf{EDIT:}$
I saw two definitions of width, so I'll put both definitions.
$\textbf{First definition of width (in terms of support function)}$: the width of a convex curve $\alpha$ in the direction $(\cos \theta, \sin \theta)$ is given by $w(\theta) = \varrho(\theta) + \varrho(\theta + \pi)$, where $\theta$ is the angle oriented positively between the normal unit vector of the curve and $e_1$ and $\varrho$ is the support function of the curve.
For the second definition of width that I see, there are some illustrations motivating the definition:
$\textbf{Second definition of width}$: for each $v \in \mathbb{S^1}$, consider the function $h$ defined by
$$h(v) := \max_{a \leq s \leq b}\langle \alpha(s), v \rangle$$
In terms of $h$, we write the width of $\alpha$ in direction $v$ as
$$\textit{larg}_v(\alpha) := h(v) + h(-v)$$
The diameter of a curve is the maximum distance between two points on trace of the curve $\alpha$.
I stated earlier that "the diameter is the maximum of width of a curve" because was proved in the book where I saw the second definition of width the following proposition:
$\textbf{Proposition}$: For every regular and closed curve $\alpha: [a,b] \longrightarrow \mathbb{R}^2$, the diameter $D$ of $\alpha$ is given by
$$D = \max_{v \in \mathbb{S}^1} \textit{larg}_v(\alpha)$$
Let $K$ be the convex body bounded by the given curve. It is clear $K$ has same diameter $D$ as its boundary.
For any $\theta \in \mathbb{R}$, let $w(\theta)$ be the width of the shadow if you project $K$ orthogonally to a line with tangent pointing in direction of $(\cos\theta,\sin\theta)$. Define the "minimum width" of the curve as
$$w_{min} \stackrel{def}{=} \min\big\{\; w(\theta) : \theta \in \mathbb{R} \;\big\}$$
WOLOG, consider the case $w_{min}$ is achieved at $\theta = 0$. i.e The projection is most narrow when the line to project is along the $x$-axis. Translate $K$ so that it is sandwiched between the line $x = 0$ and the line $x = w_{min}$. For each $x \in [0,w_{min}]$, let $h(x)$ be the length of the line segment $\{ (x,y ) : (x,y) \in K \}$. Since any such line segment is a subset of $K$, we have $h(x) \le D$. "Summing" contribution from all these line segments, we obtain:
$$\verb/Area/(K) = \int_0^{w_{min}} h(x) dx \le \int_0^{w_{min}} D dx = w_{min} D$$