That's not quite true. It's the path that requires only the gravitational force to travel.

In general relativity the gravitational force is interpreted to be an inertial force.

I agree that it is useful for calculation to take gravity as a (inertial) force. And that is because we all live in a non inertial frame of reference. I am now sitting in a couch between two opposite forces: gravity and the normal force upwards. The sum is zero, and I am not accelerating.

Another way to analyse the situation is to note that my inertial behaviour would be a free fall. On one hand, it is against our understanding of inertia as a non acelerated movement. On the other hand it is consistent with the notion of inertia as the type of movement in the absence of forces.

Generally speaking, all geodesical paths in the spacetime are inertial movements if we understand inertia as absence of forces. By the way, my path in the spacetime, being in this couch, is not a geodesical one, so it is not inertial.

In the first place we're talking not about the shortest path but an extremal path. And the magnitude of the spacetime interval between two points on a worldline in spacetime is not determined by the Euclidean metric but by the Minkowski metric. E.g. consider a photon which travels in flat spacetime between point A and point B. Suppose the spatial distance between A and B is d. The magnitude of the spacetime interval is 0. That's an example of how different Euclidean geometry is from Minkowski geometry.

I was answering a remark from yor_on about curved space, not spacetime.

In a flat spacetime, the Minkovski geometry applies, and the geodesicals are straight lines. that correlates well to uniform retinear inertia movement.

In our case, over the earth surface, the Schwartzchild metrics (neglegeting the influence of sun, moon and other planets) applies, and the geodesicals are curved lines in the spacetime. That correlates well to orbital movements being regarded as inertial ones, and being "at rest" as non inertial.