Let $\alpha :[0, l] \rightarrow \mathbb{R}^{2}$ parameterize a simple closed curve by arc length. Suppose that there is a constant $R$ such that the curvature of $\alpha$ satisfies $0<\kappa_{\sigma}(s) \leq R$. Prove that $\text{length}({\alpha}) \geq \frac{2 \pi}{R}.$
Hint: Note that the hypothesis say that the curve is curved less than the circle of radius $\frac{1}{R},$ then its length is greater than the length of the circle.
That's the problem. I was thinking about using the theorem which says that the length of a curve C in $\mathbb{R}^2$ has length $$\frac{1}{4}\int_{0}^{2\pi}length(P_\sigma(C))d\sigma,$$ where $P_\sigma(C)$ is the projection of the curve. The problem is that I don't know how to get a relation between the curvature and the projection. Maybe I am completely wrong and there is another way to solve the problem. Any ideas?
Here is a different solution: By the Hopf Umlaufsatz, $\int_C\kappa\,ds = 2\pi$. Now use $\kappa\le R$ to get $2\pi \le R\,\text{length}(\alpha)$.