Can you perturb an inscribed polytope so all its edges grow? Consider the family of convex simplicial polytopes with vertices in the unit sphere of $\mathbb{R}^n$ which have the origin as an interior point.
My question is the following:

Let $P, P'$ be two non-congruent combinatorially identical polytopes from the above family, with
vertices of each polytope labelled so that the face lattices of the
two polytopes are identical. Is it possible that $\|a-b\|\leq\|a'-b'\|$ whenever
edge $\{a,b\}$ in $P$ corresponds to edge $\{a',b'\}$ in $P'$?
In other words, can you perturb one such polytope to another making all the edges grow?

This is impossible in $\mathbb{R}^2$: Take an inscribed polygon from the family and perturb it so that it remains in the family and has the same combinatorial structure. If all the edges grow, then all the central angles grow, which would make those central angles add up to a value larger than $2\pi$. Therefore in a perturbed polygon, if some edges grow then other edges must contract.
Does this idea transfer to dimension $n=3$, or more generally to $n\geq3$? And, if yes, is there a reference?
 A: OK, let me address the case of a simplex.
In fact, it follows from the `dual Kneser--Poulsen'conjecture, as stated, e.g., in this nice paper. A good thing is that the simplex has only $n+1$ vertices, and in such partial case, according to that paper, the conjecture has been established by Gromov. To reach the desired result, apply the conjecture to the balls of unit radius centered at the vertices of both simplices. (One still needs to check that the intersection strictly changes, but that is doable).
Anyway, here is a direct proof.
Let $u_0,\dots,u_n$ be the vertices of the first simplex (hence unit vectors), and let $v_0,\dots,v_n$ be the vertices of the second one, with $\|u_i-u_j\|\leq\|v_i-v_j\|$ for all $i$ and $j$, where at least one inequality is strict.
Choose the positive $\alpha_i$ such that
$$
  \sum_i\alpha_i=1 \quad\text{and}\quad \sum_i \alpha_iu_i=0;
$$
these are the barycentric coordinates of $0$ in the first simplex. Take the point $p$ with the same barycentric coordinates in the second, i.e.,
$$
  p=\sum_i \alpha_iv_i.
$$
Clearly, $p$ lies inside the second simplex.
Notice that $\langle v_i,v_j\rangle\leq\langle u_i,u_j\rangle$. Therefore, for all $j$ we have
$$
  \langle v_j,v_j-p\rangle
  =-\sum_{i\neq j}\alpha_i\langle v_j,v_i\rangle
    +(1-\alpha_j)\langle v_j,v_j\rangle
  \geq -\sum_{i\neq j}\alpha_i\langle u_j,u_i\rangle
    +(1-\alpha_j)\langle u_j,u_j\rangle
  =\langle u_j,u_j\rangle=1,
$$
where sometimes an inequality is strict. Hence $\|v_j-p\|\geq 1$, with sometimes strict inequality; in particular, $p\neq 0$.
This shows that all the $v_i$ lie on the unit sphere centered at $0$, but outside the (open) unit ball centered at $p$. This shows that they all (along with $0$( are on the same side of the hyperplane equidistant from $0$ and $p$. Hence the second simplex cannot contain $p$ --- a contradiction.
