Physical intuition for the Geodesic equation derivation via Action Principle The most commom derivation I've seen of the geodesic equation of a massive particle is by the use of the Variational Principle. My problem is that I can't realize what the meaning of find a spacetime path (the geodesic) such that the proper time is extremized. (If the signature is $(+,-,-,-)$ it should be a maximum as some textbooks say.)
I understood that the action integral must be proportional to the line element $ds$ because we need that all the observers compute the same value of action to obtain the same equations of motion.
What I don't understand is the physical meaning of finding a maximum proper time instead a minimum, and what physical implications it leads to. How can I conclude that what I need to find a geodesic equation is maximize proper time of the massive particle? If possible, make an analogy with the Minkowski Space.
 A: *

*Let us for simplicity consider Minkowski space although the generalization to curved spacetime is straightforward. Lorentz invariance suggests that the Lagrangian one-form for a massive point particle should be
$$\mathbb{L}~=~ f(\dot{x}^2)\mathrm{d}\lambda, \qquad \dot{x}^2~:=~\eta_{\mu\nu} \dot{x}^{\mu}\dot{x}^{\nu}~>~0, \qquad \dot{x}^{\mu}~:=~\frac{dx^{\mu}}{d\lambda}, \qquad x^0~\equiv~ct, \tag{1}$$
for some function $f$. Here $\lambda$ denotes a world-line (WL) parameter, and a dot denotes differentiation wrt. $\lambda$. We chose Minkowski signature $(+,-,-,-)$ so that timelike vectors have positive length. 

*WL reparametrization invariance implies that the function $$f~\propto ~\sqrt{\cdot}\tag{2}$$ is proportional to a square root.

*In other words, the variational principle finds stationary paths for timelike arc-length (or equivalently, proper time multiplied with $c$):
$$c\tau~=~\int_{\lambda_i}^{\lambda_f}\!\sqrt{\dot{x}^2}\mathrm{d}\lambda. \tag{3}$$
The corresponding Euler-Lagrange (EL) equations are the geodesic equations. In Minkowski space the geodesics are just straight lines.

*Let us impose boundary conditions (BCs)
$$ x(\lambda_i)~=~x_i\qquad\text{and}\qquad x(\lambda_f)~=~x_f. \tag{4}$$ 
By changing coordinate system, we may assume that ${\bf x}_i={\bf x}_f.$
Finally let us choose static gauge $\lambda=t$. 

*Then eq. (3) becomes 
$$c\tau~=~\int_{t_i}^{t_f}\!\sqrt{c^2-\dot{\bf x}^2}\mathrm{d}t. \tag{5}$$
Eq. (5) is clearly maximal for $\dot{\bf x}={\bf 0}$, i.e. a particle at rest, which is also what a free particle would do with the given BCs. This answers OP's question.
