I am self studying curves and came across this problem:
Let $\alpha: I \to R^2$ be a simple smooth closed plane curve.
i) If the curve lies inside a circle of radius $r$, show there is a point at which the curvature $|k(s)|\geq1/r$.
ii) If $0<|k(s)|\leq c$ for some constant $c$. Then show that the length of the curve is bounded below by $\frac {2 \pi}{c}$.
Any help would be appreciated, this looks intuitively obvious to me as the curvature of a circle is 1/r so there must be a point on that curve with greater curvature as it is inside the circle, but how to prove it mathematically? Thanks in advance.
Edit: I have tried to use the fact that $<\alpha(s),\alpha(s)> \leq r^2$ and do some algebra or use serret frenet out of it but to no avail. (by < , > i mean dot product)