My question is ( an exercise of Riemannian geometry ):

Let $M$ be a complete Riemannian manifold, suppose there exists $c>0$ such that the sectional curvature of $M$ is greater than or equal to $c$. If $\gamma$ is a closed geodesic on $M$, prove that $\forall p\in M$, the distance between $p$ and $\gamma$ (namely, $d(p,\gamma)$) is less than or equal to $\frac{\pi}{2\sqrt{c}}$.

By Bonnet-Myers theorem, the diameter of $M$ is less than or equal to $\frac{\pi}{\sqrt{c}}$.

If $M$ is a $n$-dimensional ball of radius $\frac1{\sqrt{c}}$ ( denoted by $B^n(\frac1{\sqrt{c}})$), then $M$ has constant curvature $c$ and the closed geodesic on $M$ must be the great circle, the conclusion is clearly valid.

In general, the condition on sectional curvature (rather than Ricci curvature) leads me to the Rauch comparison theorem, but I don't know how to use it.