Theorem (??) derived in this MO-post from Schoenberg's theorem yeilds a "bipartite" characterization of metric spaces that admit an isometric embedding into a Hilbert space. This Theorem (??) implies the following necessary condition of bi-Lipschitz embeddability into a Hilbert space.

**Theorem.** If a metric space $X$ admits a bi-Lipschitz embedding to a Hilbert space, then there exists a positive real constant $L$ such that the inequality
$$\sum_{i<j}d^2(x_i^+,x_j^+)+\sum_{i<j}d^2(x_i^-,x_j^-)\le L\cdot \sum_{i,j}d^2(x_i^+,x_j^-)\quad\quad\quad(*)$$holds for any points $x_1^+,\dots,x_n^+,x_1^-,\dots,x_n^-\in X$.

**Remark.** The inequality $(*)$ in Theorem is not sufficient for the existence of a bi-Lipschitz embeding of $X$ to a Hilbert space. Indeed, the triangle inequality implies that for every metric $d$ on set $X$ the metric $\sqrt{d}$ satisfies the inequality $(*)$ with $L=2$. It is known that the unit ball $(B,d)$ of the Banach space $c_0$ does not admit a uniform embedding into a Hilbert space. Then the metric space $(B,\sqrt{d})$ satisfies the inequality $(*)$ (with $L=2$ but admits no bi-Lipschitz embedding to a Hilbert space.

Problem 1.Is there any nice geometric condition (or inequality) which is necessary and sufficient for the existence of a bi-Lipschitz embedding of a given metric space to a Hilbert space?

Problem 2.Is there a positive $\varepsilon$ such that each metric space $X$ satifying the inequality $(*)$ for $L=1+\varepsilon$ admits a bi-Lipschitz embedding to a Hilbert space?