Let $x,y,z \in \mathbb{R}^n$ and define the euclidean angle between two vectors as $$ a(x,y) := \arccos\left(\frac{x^\top y}{\|x\|_2\|y\|_2}\right). $$ Assuming $\|x\| = \|y\| = \|z\| = 1$ for a general norm $\|\cdot\|$ (i.e., not necessarily the $\ell_2$ norm $\|\cdot\|_2$), I would like to show that $$ a(x,y) \leq a(x,z) \iff \|x-y\|_2 \leq \|x-z\|_2. \tag{1} $$
Geometrically, this makes sense to me. However, I have been trying to prove it using the identity $2x^\top y = \|x\|_2^2 + \|y\|_2^2 - \|x-y\|_2^2$, the Cauchy–Schwarz inequality and equivalence of norms, without success.
My questions are:
- Is (1) true?
- If not, does it hold for some special case other than $\|\cdot\| = \|\cdot\|_2$? Perhaps if we define the angle using a general norm as well instead of the $\ell_2$ norm? Or if the inequality on the right hand side of (1) used a general norm as well?
Edit 1: fixed typo in the definition of angle.
First question: No.
Take the max norm on $\mathbb R^2$, for example. Your conjecture implies that $a(x,y)=a(x,z)\iff\lVert x-y\rVert_2=\lVert x-z\rVert_2$, provided that $\lVert x\rVert_\infty=\lVert y\rVert_\infty=\lVert z\rVert_\infty=1$.
Let $x=(1,1/2)$, $y=(1,0)$, and $z=(1,1)$. Clearly the Euclidean distances are both $1/2$, but the angles are not equal: $\arccos(2/\sqrt5)\neq\arccos(3/\sqrt{10})$.