Does closed Alexandrov space admit a bi-Lipschitz embedding into $\mathbb R^N$?

See Distance embedding (27.5) in our book

In my preprint "Bi-Lipschitz embeddings of SRA-free spaces into Euclidean spaces" (https://arxiv.org/abs/1906.02477) I prove the quantitative version of this statement.

Theorem 2. For $n \in \mathbb{N}$, $k < 0$ and $R > 0$ there exist $D > 0$ and $N \in \mathbb{N}$ satisfying the following. For every $n$-dimensional Alexandrov space of curvature $\ge k$ and every $x \in X$ there exists an embedding $\phi:B_R(x) \rightarrow \mathbb{N}^N$ which bi-Lipschitz distortion does not exceed $D$.

For $n \in N$ there exist $D_0 > 0$ and $N_0 \in \mathbb{N}$ such that every $n$-dimensional Alexandrov space of non-negative curvature allows an embedding into $\mathbb{N}^{N_0}$ which bi-Lipschitz distortion does not exceed $D_0$.