Preliminaries
Suppose that $\Gamma$ is a simple (sufficiently smooth) closed curve in $\mathbb{R}^2$ with length $L$ and enclosing an area $A$. Suppose furthermore that $\gamma:[0,L]\rightarrow\mathbb{R}^2$ is a parametrization-by-arclength for $\Gamma$ so that $|\gamma '(s)|=1$ for every s, i.e.
$$L = \int_0^{L} |\gamma'(s)| $$
The total (absolute) curvature can be defined by
$$C = \int_0^{L} |\gamma''(s)| = \int_0^{L} \kappa(s)$$
It can be shown that $C\ge 2\pi$. Equality holds iff $\Gamma$ is convex.
$$4\pi A\le L^2$$
(equality holds iff $\Gamma$ is a circle) can be stated using another derivative of $\gamma$
$$\Omega = \int_0^{L} ||\gamma''(s)|'| = \int_0^{L} |\kappa'(s)|$$
A closed curve $\Gamma$ is a circle $\mathsf{O}$ iff $\Omega = 0$. (Edit: Pietro correctly pointed out that the initial formula $\int_0^{L} |\gamma'''(s)|$ would not make sense.) This implies
$$\left(( L_{\Gamma} = L_{\mathsf{O}}) \wedge (C_{\Gamma} = C_{\mathsf{O}}) \wedge (\Omega_{\Gamma} \ge \Omega_{\mathsf{O}})\right) \rightarrow (A_{\Gamma} \le A_{\mathsf{O}})$$
for all closed curves $\Gamma$ and circles $\mathsf{O}$.
Question
I wonder if and how it could be shown that a generalization of the last proposition holds:
$$ \left((L_{\Gamma} = L_{\Gamma'}) \wedge (C_{\Gamma} = C_{\Gamma'}) \wedge (\Omega_{\Gamma} \ge \Omega_{\Gamma'})\right) \rightarrow (A_{\Gamma} \le A_{\Gamma'}) $$
for all closed curves $\Gamma, \Gamma'$, i.e. of two curves of the same length and the same absolute curvature the one with greater change rate of curvature has the smaller area.
How would this – probably using Fourier analysis – be shown for ellipses with the same perimeter?
