Given a closed curve in the plane $\mathbb{R}^2$, it is well known that $L^2 \geq 4\pi A$ where $L$ is the length of the curve and $A$ is the area of the interior of the curve.

For a *simple* closed curve $\gamma$, the stronger inequality due to Bonnesen holds:
$L^2 - 4\pi A \geq \pi^2 (R_{out}- R_{in})^2$,
where, setting $\Omega =$ Int $(\gamma) $, $ R_{in}$ and $R_{out}$ denote the inner and outer radii of the sets:

$R_{in} = \sup_{B_r \subset \Omega} r$

$R_{out} = \inf_{\Omega \subset B_R} R$

**Question:** Does this inequality continue to hold **without** the assumption that the curve is **simple**? In particular, does it hold for any *connected, rectifiable set*?

If one adds components to the interior of a simple curve, it is clear that this increases the isoperimetric defecit. However while $R_{out}$ will remain unchanged, $R_{in}$ will necessarily decrease, making it not immediately clear that the inequality would continue to hold.

**Update:** I just realized this is *false*. Take a countable union of points in the interior of the unit ball. Around the kth point, make a circle of radius $\epsilon/2^k$. Distribute the points well enough and $R_{in}$ can be made arbitrarily small. Then if you make $\epsilon$ small enough, you don't change the length or area too much and so the desired inequality will be violated.