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A metric space $(X,d)$ is isometrically homogeneous if its isometry group acts transitively on points, i.e., for every $x,y \in X$ there is an isometry $\varphi:X\to X$ with $\varphi(x) = y$. I'd lik …

Here's an answer if you make a further assumption on your metric space: if $M$ is of strictly negative type, then it has $n-1$ negative eigenvalues, according to Lemma 3.6 of this paper. This conditi …

The second example in the original post generalizes a lot. Let $d_i$ be finitely or countably many pseudometrics (it's possible for $d_i(x,y)=0$ even if $x\neq y$) for $i\ge 1$, and assume $d_1$ is a …

I assume you also want your compact sets to have non-empty interior, hence positive volume. The literature mostly deals with the related Banach-Mazur metric $d_{BM}(A,B)$, in which it is assumed that …

There are many results, and an active research industry, along these lines. In general the Euclidean ball is the best-behaved convex body in this respect, and just how similar an arbitrary convex bod …

Q1: Most often it is the maximal volume ellipsoid contained in $K$. Q2: (a) John's theorem implies that $E^+ \subseteq d E^-$ in general, and $E^+ \subseteq \sqrt{d} E^-$ if $K$ is centrally symmetri …

It is a classical fact that if $(x_1,\ldots,x_n)$ is a random vector uniformly distributed on the sphere $S^{n-1} \subseteq \mathbb{R}^n$, then the random vector $(x_1,\ldots,x_{n-2})$ is uniformly di …

The usual approach is to generate $n+1$ i.i.d. mean zero Gaussian random variables $X_1, \dotsc, X_{n+1}$ to get a random point $X$ in $(n+1)$-space with rotationally invariant distribution and normal …

No, it is not known that the minimum is unique, but it is believed to be. In fact, this paper by Kim and Reisner proves that the simplex is (modulo linear equivalence) a strict local minimum; thus th …

A classical result along these lines is Bonnesen's inequality, which states $$ L^2 - 4\pi A \ge \pi^2 (r_{out} - r_{in})^2, $$ where $L$ is the length and $A$ is the enclosed area of a simple planar c …

A different symmetrization-based proof is given in this review article by Schechtman (pp. 7-8); see the previous page for references.

I think "compact" can be even weakened here to "separable and complete" (and regarding your first guess, total boundedness is essentially used to prove that compact implies separable). Here's a sketc …