Manifolds with geodesics which minimize length globally I am interested in complete Riemannian manifolds whose geodesics minimize length globally. Such manifolds must be non-compact (otherwise there is always a self-intersecting geodesic)
However, I suspect this property is much more restrictive.
Question: Assume $(M,g)$ is complete and has this property. 
Must $M$ be simply connected? The exponential map has to be a diffeomorphism on all $T_pM$?
Does the sectional curvature has to be non-positive? Are there unique geodesics between any two points?
Update: From John Ma's answer, it turns out that the $exp_p$ is a diffeomorphism, and in particular there are unique geodesics between any two points. 
I would still like to know if anything intelligent can be said about the curvature though. 
My guess is that it does not has to be non-positive  everywhere, but only 'mostly everywhere' in some sense. (i.e I can imagine a surface with small regions of positive curvature which doesn't violate our condition )

I am looking in general for necessary and sufficient conditions (topological\curvature constraints) for this property to hold.
One sufficient conditions is provided by Hadamard's Theorem.
 A: Partial answer
Let $M$ be a complete Riemannian manifold. Then every two points can be joined by a length minimizing geodesic ($\gamma$ is called length minimizing between $p$, $q$ if for any piece-wise smooth curves $\eta$ joining $p$ and $q$ we have $L(\eta)\ge L(\gamma)$). The following are equivalent:


*

*All geodesics in $M$ are length-minimizing. 

*All points in $M$ are joined by a unique geodesic.

*The exponential map $\exp_p : T_pM \to M$ is a diffeomorphism for all $p \in M$. 
$(1)\Rightarrow (2)$: If not, let $p, q\in M$ and $\gamma_1, \gamma_2 : [0,1] \to M$ be two geodesics joining $p, q$. Then both $\gamma_1, \gamma_2$ cannot be length minimizing when $t>1$. 
$(2)\Rightarrow (1)$ is obvious, since length minimizing curve must be a geodesic and every two points can be joined by at least one geodesic as $M$ is complete. 
$(2)\Rightarrow (3)$: By completeness, $\exp_p$ is surjective. If it is not injective, then there are two geodesics which starts at $p$ and interest at some points. Indeed it has to be a diffeomorphism. If not, then there are two points that are conjugate to each other. Thus there are (yet another) two points $p, q$ so that they are joined by a geodesic $\gamma$ which is not length minimizing. By completeness, $p$ and $q$ are also joined by another geodesic $\eta$, thus it contradicts $(2)$. 
$(3)\Rightarrow (2)$ is also obvious. 
So there are huge topological constraint given by $(3)$. Not completely sure about the curvature constraint, except the one you mentioned. 
A: I am just elaborating on some points related to John's answer:
$(1)\Rightarrow (2)$: By completeness, every two points are joined by a geodesic. Assume there exists $p,q \in M, \gamma_1, \gamma_2 : [0,L] \to M$ two geodesics (parametrized by arclength) joining $p,q$. Then both $\gamma_1,\gamma_2$ cannot be length minimizing when $t>L$:
Assume $\gamma_1$ is minimizing between $p=\gamma_1(0),\gamma_1(L+\epsilon)$. 
Define a path $\beta=\gamma_1|_{[L,L+\epsilon]} *\gamma_2|_{[0,L]}$. 
$L(\beta)=L+\epsilon=L(\gamma_1|_{[0,L+\epsilon]})=d(p,\gamma_1(L+\epsilon))$
So $\beta$ is length minimizing between $p,\gamma_1(L+\epsilon)$ so it's  a geodesic*, and in particular smooth. But this implies $\dot \gamma_1(L)=\dot \gamma_2(L)$, so by uniqueness $\gamma_1=\gamma_2$.
$(2) \Rightarrow (1)$: Assume there is a geodesic between $p,q$ which is not minimizing. Hopf-Rinow theorem implies there is a minimizing geodesic between $p,q$, so there is more than one which is a contradiction.
Note: $(3) \Rightarrow (1)$ is also immediate, since we know geodesics minimize distance from a given point $p$ as long as they stay in a normal ball around $p$. (See Do-carmo's Riemannian geometry, proposition 3.6). If $exp$ is a diffeomorphism, then all $M$ can considered as a normal ball.

*See Do-carmo's Riemannian geometry, proposition 3.9
