I'm trying to prove Osserman's third lemma in his proof of the Four Vertex Theorem:
Lemma 3. Let a smooth oriented unit speed curve $\gamma$ have the same unit tangent vector $\vec{t}$ at a point $P$ as a positively oriented circle $C$ of radius $R$. Let $\kappa$ be the curvature of $\gamma$. Then if $\kappa(P) > 1/R$, a neighborhood of $P$ on $\gamma$ lies inside $C$, while if $\kappa(P) < 1/R$, a neighborhood of $P$ on $\gamma$ lies outside $C$.
It seems intuitive that that if the curvature of $\gamma$ is less than the curvature of the circle, then a portion of the $\gamma$ near $P$ should lie outside of the circle, while if the curvature of $\gamma$ is greater than the curvature of the circle, then a portion of the $\gamma$ near $P$ should lie inside of the circle.
I'm having trouble finding a rigorous solution, though, and I haven't been able to find anything relevant online.
My attempt at a proof
Consider the case where $\kappa(P) > 1/R$ at $P$. Now the osculating circles of $\gamma$ must have radius $\frac{1}{\kappa}<R$. By assumption, $\gamma$ and $C$ have the same unit normal vector at $P$, so the osculating circles of $\gamma$ on lie inside $C$.
If $\kappa' \neq 0$ on $P$, we can show algebraically that the osculating circle of $\gamma$ intersects with $\gamma$ on a neighborhood of $P$, and we are done. If $\kappa$ is constant on a neighborhood of $P$, then $\gamma$ must coincide with its osculating circle on that neighborhood, and we are also done. Otherwise, since $\gamma$ is smooth, there is a point infinitesmally close to $P$ with $\kappa' \neq 0$ and $\kappa(P) > 1/R$, so we have the same result.
The case where $\kappa(P) < 1/R$ is extremely similar.
This feels very hand-wavy and probably incorrect -- especially in the case where $\kappa' = 0$. (I found that requirement for the intersection with the osculating circle from an answer on here). Any advice would be appreciated!
