Let $X$ be an $n$dim Alexandrov space with curvature $\geq k$. Does $X$ satisfy a reverse doubling condition? That is, does there exist a constant $C>1$, s.t., for any $x\in X$, $0<r<\mathrm{diam}(X)/2$, $\mathrm{vol}(B_x(2r))\geq C\cdot\mathrm{vol}(B_x(r))$?

1$\begingroup$ I think you have to bound $r$ from above. Otherwise consider $S^n$ with the standard metric and put $r=\pi$. Then the two balls are the same (except for the antipode). Even worse: flat $\mathbb R^n$ has quotients of arbitrarily small diameter. But maybe you stand a chance if you demand that $r\le\frac12\mathrm{diam}(X)$. $\endgroup$ – Sebastian Goette Jan 6 '16 at 10:13

$\begingroup$ @SebastianGoette Thanks for you suggestion. $\endgroup$ – Shman Jan 6 '16 at 15:19
The answer is no. In fact, it is not even true for Riemannian manifolds.
Consider a flat Riemannian manifold $(M,g)$ of dimension $m$ with diameter $d$ realized by the points $x$ and $y$. Let the volume be $Cd^m$. Let $M_n$ denote the Riemannian manifold $(M,n^2 g)$, where $n$ is a positive integer.
Attach a cylinder of radius $\delta$, where $\delta \ll d$ at the point $y \in M_n$. (More formally, remove $y$ from $M_n$, and change the metric so that it is a product outside a compact subset.) Do this so that the curvature remains $\geq 1$. The "bending" required to attach the cylinder can be the same for every $n$, because the metric is flat.
Now in $M_n$ we have $\mathrm{vol}(B_x(nd)) = C(nd)^m + \epsilon_1$ and $\mathrm{vol}(B_x(2nd)) = C(nd)^m + C'nd\delta^{m1} + \epsilon_2$, where $\epsilon_1$ and $\epsilon_2$ are determined by the bending, and so are independent of $n$, and $C'=C'(m)$.
Then the ratio of the volumes tends to 1 as $n \to \infty$. You can attach copies of $M_n$ for every $n$ to the Euclidean plane to get a Riemannian manifold of curvature $\geq 1$ which does not satisfy the condition.