Let $(M,g)$ be a Riemannian manifold. As part of an exercise, I have proved the following two facts:
For every $m\in M$ there exists $\varepsilon >0$ and an open neighborhood $U$ of $m$ such that: for every $x$ and $y$ in $U$ there is $v\in T_x M$ with $||v||<\varepsilon$ and such that $\textrm{exp }(x,v)=y.$ I have proved this by using the inverse function theorem.
If $\gamma$ is a non-radial geodesic contained in some geodesic ball $\textrm{exp}_m (B(\varepsilon))$ then $r\circ \gamma$ attains its maximum values at the endpoints (cannot have strict maximums). To prove this I have used the geodesic equation locally.
Now I have to show as a consequence of the previous facts that every point has a geodesically convex neighborhood, i.e. that for every pair of points there is a lengh minimizing geodesic joining them, and such that it is contained in the neighborhood for every time. It seems it should be easy to solve regarding the previous facts, but I am not able to prove it...
Indeed, this is easy given 1, 2.
Choose $\delta <\epsilon /2$ so that $B_\delta (m) \subset U$. Let $x, y\in B_\delta (m)$. Then by 1 there is $v\in T_xM$ so that $\exp_x(v) = y$. The geodesic $t\mapsto \exp_x(tv)$ has length $\|v\| <\epsilon$. Note that this geodesic is inside $B_{\epsilon}(m)$ (or it has length $\ge \epsilon$), by 2 we have that
$$ r(\gamma(t) \le \max\{r(\gamma(0)), r(\gamma(1))\}<\delta.$$
Thus $\gamma(t) \subset B_\delta(m)$ for all $t$. This proves that $B_\delta(m)$ is geodesically convex.