Toponogov's theorem in the Alexandrov space with respect to a compact set Let $X$ be an Alexandrov space with curvature $\ge 0$ and $A \subset X$ a compact set. Suppose $p,q\in X$ satisfying $|Ap|=|Aq|=r$ and $p_1,q_1$ are points on the geodesics $Ap,Aq$ respectively such that $|Ap_1|=|Aq_1|=s<r$. Can we prove
$$ 
|p_1q_1| \ge \frac{s}{r}|pq|?
$$
 A: Yes it is true. It can be proved by applying distance estimates to the gradient flow of the function $f=\mathrm{dist}_A^2/2$. See our draft.
A: Yes. It is also true (with different distortion constant) for Alexandrov space with curvature $\ge k$.
We define
$$
\rho_k(x)=
\begin{cases}
\frac{1}{k}(1-\cos(x\sqrt{k}))\quad &\text{if}\quad k>0,\\
\frac{x^2}{2},\quad &\text{if}\quad k=0,\\
\frac{1}{k}(1-\cosh(x\sqrt{-k}))\quad &\text{if}\quad k<0.\\
\end{cases}
$$
Then it follows from the Toponogov's theorem that for any $p \in X$,
$$
\rho_k(\text{dist}_p)''\le 1-k\rho_k(\text{dist}_p).
$$
By taking the minimum for all $p \in A$, we also have
$$
f''\le 1-kf,
$$
where $f:=\rho_k(\text{dist}_A)$.
Let $\gamma_1(t)$ and $\gamma_2(t)$ be the geodesics of $Ap$ and $Aq$ respectively. It is not hard to see that $\tilde \gamma_1(z):=\gamma_1(t(z))$ and $\tilde \gamma_2(z):=\gamma_2(t(z))$ are the gradient flows of $f$ starting from $p_1$ and $q_1$ respectively. Here the new parameter $z$ is determined by
$$
dz=\frac{dt}{\rho_k'(t)} \quad \text{and} \quad t(0)=s.
$$
If we set $l(t):=|\gamma_1(t)\gamma_2(t)|$, then it follows from the distance estimate of the gradient flow (see Lemma 1.3.3 in this paper) that
$$
(\log l(t))' \le \frac{1-k\rho_k(t)}{\rho_k'(t)}.
$$
From integration, we obtain
$$
|pq|=l(r) \le \sigma_k(r,s) l(s)=\sigma_k(r,s)|p_1q_1|,
$$
where
$$
\sigma_k(r,s)=
\begin{cases}
\frac{\sin(r\sqrt k)}{\sin(s\sqrt k)}\quad &\text{if}\quad k>0,\\
\frac{r}{s},\quad &\text{if}\quad k=0,\\
\frac{\sinh(r\sqrt{-k})}{\sinh(s\sqrt{-k})}\quad &\text{if}\quad k<0.\\
\end{cases}
$$
