A simple proof for angle inequality in inner product spaces

Question

I am looking for a smiple proof for the following fact: Let $u,v,w$ be vectors in an inner product space $V$. Then it holds:

$\theta (u, v)≤\theta(u, w) + \theta(w, v)$

(Of course if they are all in a single plane and $w$ is "between" $u,v$ there is an equality).

So far I have only found proofs which seem a bit too complicated then necessary. (For instance here).

Answer 1

This can be thought of as the triangle inequality applied to a "geodesic" metric on the sphere. In particular: given unit-vectors $u$ and $v$, $\theta(u,v)$ is the minimal distance from $u$ to $v$ along a curve on the surface of the unit sphere.

If we define the angle via this minimum (or show that the angle is equal to this minimum), then of course the inequality holds: the minimum length of all arcs from $u$ to $v$ is certainly at most the minimum length of all arcs from $u$ to $v$ that pass through $w$.

I would like to think that establishing the equivalence of these notions would be a routine application of the calculus of variations. Then again, I have not at all tried this for myself. I also have no idea whether this idea would extend to infinite-dimensional spaces.