TITLE: Are hyperbolic spaces actually better for embedding trees than Euclidean spaces?
QUESTION [10 upvotes]: There is a folklore in the empirical computer-science literature that, given a tree $(X,d)$, one can find a bi-Lipschitz embedding into a hyperbolic space $\mathbb{H}^n$ and that $n$ is "much smaller" than the smallest dimension of a Euclidean space in which $(X,d)$ can be bi-Lipschitz embedded with similar distortion.
Question A: Is there any theoretical grounding to this claim?  Namely, can one prove that $(X,d)$ (where $\# X = n\in\mathbb{N}_+$) admits a bi-Lipschitz embedding into some $\mathbb{H}^n$ with:

distortion strictly less that $O(\log(n))$
$n<O(\log^2n)$?

! Edit - (Following Discussion of YCor, WillSawin, and TomTheQuant): What can be said if $s=1$ in Equation (1)?
Question B (Converse):  For every $n\in \mathbb{N}_+$ and every $D>0$ does there exist a finite metric space $(X,d)$, which don't admit a bi-Lipschitz (resp. possibly uniform embedding) into $\mathbb{H}^n$ with distortion at-most $D$?
Relevant Definition (For completeness)
A bi-Lipschitz embedding $f:X\rightarrow \mathbb{H}^n$ of a metric space $(X,d)$ into $\mathbb{H}^n$ with distortion $D>0$ is a Lipschitz homeomorphism $f:X\rightarrow \mathbb{H}^n$ Lipschitz inverse $f^{-1}$ such that there is some $s>0$ satisfying
$$
sd(x_1,x_2) \leq d_{\mathbb{H}^n}(f(x_1),f(x_2)) \leq sDd(x_1,x_2)
\qquad (1)
$$
for every $x_1,x_2\in X$.  Here, $d_{\mathbb{H}^n}$ is the usual geodesic distance on the $n$-dimensional hyperbolic space.
Some Relevant posts:
Hyperbolic embeddings

References to embedding into hyperbolic spaces
Representability of finite metric spaces

Flat Embeddings

Problem with embedding expanders into "flat" spaces
Characterizing finite metric spaces which embed into Euclidean space

Uniform Embeddings

Notes on coarse and uniform embeddings

REPLY [3 votes]: Here is a trivial example for question B (in the $s=1$ case):
The discrete metric space on $N$ points with distance 1 between every two distinct points has the minimal distortion of an embedding into $\mathbb H^n$ going to $\infty$  as $N$ goes to $\infty$ with respect to $n$.
Indeed, for an embedding of distortion $D$, any two points in the embedding must have distance at least $1$, so the balls of radius $1/2$ around these points must be disjoint.  But every two points have distance at most $D$, so the balls of radius $1/2$ around these points must be contained in the ball of radius $D+1/2$ around one point.
Thus, for an embedding to exist, the volume of the ball of radius $D+1/2$ must be at least $N$ times the volume of the ball of radius $1/2$.
Thus $D$ must go to $\infty$ if $N$ goes to $\infty$ with fixed $n$.