I'm trying to solve the following problem:
(1) Given a convex, closed plane curve $\alpha$ without cusps that encloses an area $A$, prove that there exists a point $t$ such that the (unsigned) curvature $k(t)$ satisfies $k(t) \geq \sqrt{\dfrac{\pi}{A}}$.
My approach thus far: assuming that $\alpha$ is differentiable enough and parametrized by arclength, we have that $k(s)=|\alpha''(s)|$ is continuous in a compact set (since the curve is closed) and thus reaches a maximum $k(s_0)=K:=\displaystyle \max_s k(s)$. If we can show that the osculating circle at $\alpha(s_0)$ lies entirely inside the area $A$, we're done since this circle has a radius $R = 1/K$ and encloses an area $A_\circ=\pi R^2 \leq A$, and then we have: $$k(s_0)=K=\sqrt{\dfrac{\pi}{A_\circ}} \geq \sqrt{\dfrac{\pi}{A}}.$$
Therefore, problem (1) boils down to the following:
(2) Given a convex, closed plane curve $\alpha$ without cusps that encloses an area $A$ and reaches its maximum curvature at a point $P$, does the osculating circle at $P$ lie completely within $A$?
This intuitively feels true since $k$ determines the curve up to isometries, so in points/intervals where $k(s)=K$ the curve should be locally isometric to a section of the osculating circle, and if $k(s)<K$ the curve won't cut a tangent copy of the same circle. I also know that $k(t)=\theta'(t)$, but either way I can't seem to articulate any of this to produce a proof.
Two related questions I found on this site are:
Relations between curvature and area of simple closed plane curves. and
Osculating circle at curvature minimum point of simple closed curve encloses the curve,
which are sort of "dual" to what I want to prove, but a similar approach (e.g. involving the total curvature formula and the isoperimetric inequality) doesn't seem to work directly here.
How can I prove (2), if it is true? Is my strategy for (1) good or does it follow from a simpler statement?
Note that (1) is not necessarily true if we allow the plane curve $\alpha$ to have cusps: a convex polygon has constant curvature $0$ and encloses a non-trivial area, and a semicircle of radius $R$ has points of curvature $k(t)=0$ or $k(t)=1/R$ and encloses an area $A=\pi R^2/2$, and thus in this case we have $k(t)<\sqrt{\dfrac{\pi}{A}}$ for all $t$. It doesn't even suffice to assume non-zero curvature: imagine a convex polygon where the edges are replaced by circle sections of a large enough radius (i.e. small enough curvature). I imagine it's also easy to produce counterexamples if the curve isn't convex, which might be why the approach in the linked questions doesn't work here.