I'm sure that many readers are already familiar with the well known Bonnesen inequality in the plane for a smooth, connected curve: 


$(R\_{out} - R\_{in})^2 \leq \pi^2 (L^2 - 4\pi A),$


where $R\_{out}$ and $R\_{in}$ are the outer and inner radii respectively, $L$ is the length of the curve and $A$ is it's area. 


This form turns out to often be quite inconvenient for me, and I was wondering if there was a version which read as follows:


$R\_{out}^2 - R\_{in}^2 \leq C (L^2 - 4\pi A)$,


where now $C$ is a constant which may depend on the support of the curve or other quantities possibly. The main difference is that when $R\_{in} - R\_{out}$ is small, the second inequality I've written is *strictly stronger* if one assumes the curve is constrainted to a bounded domain. The sacrifice I make though is that I do not care what the constant is, or if it depends on the support of the curve.


This inequality seems quite reasonable but I have thus far been unable to prove it. Has anyone encountered such a version of this inequality?



Different forms of Bonnesen's strong isoperimetric inequality in the plane.
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
$$
4 s m(f) +o(s)\leq C o(s)
$$
which is clearly impossible after letting $s\to 0$. 


It works for Bonnesen's inequality as there is a favorable cancellation.


