I've been rediscovering inner product spaces recently and have developed a couple of questions about angles defined in general inner product spaces.
Consider the real vector space $V$ equipped with inner product $\langle \dot{},\dot{}\rangle : V \times V \rightarrow \mathbb{R} $ .
Take $X,Y,Z\in V$. Define the angle between $X$ and $Z$ as $$\theta_{XZ} := \cos^{-1}\left(\frac{\langle{}X,Z\rangle{}}{||X|| \dot{}||Z||}\right)$$ and define $\theta_{XY}$ and $\theta_{YZ}$ similarly. Then is it possible to prove that:
(i) (a sort of triangle inequality)
$$\theta_{XZ} \leq \theta_{XY} + \theta_{YZ}$$
(ii) in the case where $Z=Y-X$, $$\theta_{XY}+\theta_{YZ}+\theta_{XZ}=\pi$$ (i.e. the interior angles of a triangle sum to $\pi$)?
In the case $V=\mathbb{R}^2$, these results are known to secondary school children! Are they true in any real inner product space?
Thanks!
Statements (i) and (ii) are unchanged if we replace $V$ with the subspace spanned by $X$, $Y$, and $Z$. So for (i) we may assume $V$ is $3$-dimensional, and for (ii) we may assume $V$ is $2$-dimensional. Every finite-dimensional real inner product space is isomorphic to $\mathbb{R}^n$ (with the usual dot product) for some $n$, so without loss of generality we may assume $V=\mathbb{R}^3$ in (i) and $V=\mathbb{R}^2$ in (ii).
As you say, with a slight modification, (ii) is known to secondary school children. The statement should be $$ \theta_{XY} +\theta_{(-Y)(-Z)} + \theta_{(-X)Z} = \pi, $$ which is equivalent to the modification you suggest in the comments.
Statement (i) is invariant under scaling, so we may assume $X$, $Y$, and $Z$ are unit vectors in $\mathbb{R}^3$. The angle between two unit vectors is the same as the great circle distance between the corresponding points on the unit sphere, so (i) follows from the well-known fact that the shortest path between two points on a sphere is a great circle.