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I seriously doubt there is any such estimate (at least one that doesn't have the constant $C$ depend in a complicated and extremely unnatural way on the geometry of the curve).

My heuristic reasoning is as follows. Let $\sigma$ be the unit circle and $f$ be a non-zero smooth function on $\sigma$ so that $\int_{\sigma} f=0$ and $\max_\sigma f =-\min_\sigma f:=m(f)>0$.

Set $$\sigma_s=\lbrace (1+sf) \mathbf{x}(p): p\in \sigma\rbrace\subset\mathbb{R}^2$$ where here $\mathbf{x}(p)=(x_1(p),x_2(p))$ are the restrictions of the coordinate functions. For $s$ small we have we have that $\sigma_s$ is a smooth Jordan curve (and is smooth and with geometry as close to that of the unit circle as desired).

If $L_s$ is the length of $\sigma_s$, $A_s$ the area of the enclosed region and $R_s^\pm$ the in and out radius we have $$L_s=L_0+o(s)=2\pi +o(s)$$ $$A_s=A_0+o(s)=\pi+o(s)$$ $$R_s^\pm=R_0^\pm \pm sm(f) +o(s)=1\pm s m(f)+o(s)$$

If your inequality were to hold we would then have a constant $C$ so that $$2 4 s m(f) +o(s)\leq C o(s)$$ which is clearly impossible after letting $s\to 0$.

It works for Bonnesen's inequality as the linear term in the perturbation of in and out radius cancelsthere is a favorable cancellation.

1

I seriously doubt there is any such estimate (at least one that doesn't have the constant $C$ depend in a complicated and extremely unnatural way on the geometry of the curve).

My heuristic reasoning is as follows. Let $\sigma$ be the unit circle and $f$ be a non-zero function on $\sigma$ so that $\int_{\sigma} f=0$ and $\max_\sigma f =-\min_\sigma f:=m(f)>0$.

Set $$\sigma_s=\lbrace (1+sf) \mathbf{x}(p): p\in \sigma\rbrace\subset\mathbb{R}^2$$ where here $\mathbf{x}(p)=(x_1(p),x_2(p))$ are the restrictions of the coordinate functions. For $s$ small we have we have that $\sigma_s$ is a Jordan curve (and is smooth and with geometry as close to that of the unit circle as desired).

If $L_s$ is the length of $\sigma_s$, $A_s$ the area of the enclosed region and $R_s^\pm$ the in and out radius we have $$L_s=L_0+o(s)=2\pi +o(s)$$ $$A_s=A_0+o(s)=\pi+o(s)$$ $$R_s^\pm=R_0^\pm \pm sm(f) +o(s)=1\pm s m(f)+o(s)$$

If your inequality were to hold we would then have a constant $C$ so that $$2 s m(f) +o(s)\leq C o(s)$$ which is clearly impossible after letting $s\to 0$.

It works for Bonnesen's inequality as the linear term in the perturbation of in and out radius cancels.