Edit: The previous title didn't really ask the same thing as the question (sorry about that), so I've changed it. To clarify, I understand that the action isn't always a minimum. My questions are in the points 1. and 2. below.
I understand that "principle of least action" is somewhat of a misnomer, since we find that to determine the path taken by a system, we need only impose the condition that the action be stationary, i.e., that $\delta S$ should vanish to first order for small variations of the path, and this leads to the Euler-Lagrange equations.
In The Classical Theory of Fields, Landau & Lifshitz discuss the relativistic action for a free particle:
So for a free particle the action must have the form
$$S = -\alpha \int_a^b ds$$
(...) It is easy to see that $\alpha$ must be a positive quantity for all particles. In fact, as we saw [earlier], $\int_a^b ds$ has its maximum value along a straight world line; by integrating along a curved world line we can make the integral arbitrarily small. Thus the integral $\int_a^b ds$ with the positive sign cannot have a minimum; with the opposite sign it clearly has a minimum, along the straight world line.
There is also a footnote, addressed a couple of paragraphs earlier, but which is relevant:
Strictly speaking, the principle of least action asserts that the integral $S$ must be a minimum only for infinitesimal lengths of the path of integration. For paths of arbitrary length we can say only that $S$ must be an extremum, not necessarily a minimum.
I have two questions regarding this:
How is the condition "the action must be a minimum for infinitesimal displacements" formulated? I've never heard of that outside of Landau & Lifshitz's books, and in Mechanics they mention it as well but doesn't go into detail. Is this discussed a bit more somewhere?
If for the whole path the action only needs to be stationary, how can we make the argument for the negative sign? If the action had to be a minimum then it would make sense, but surely the fact that $\delta S$ = 0 isn't affected by an overall sign?