Sufficient condition for the equivalence of metric segments For a metric space $(X,d)$ and points $x,y \in X$ we define the metric segment between them as the following set:
$\left [ x,y \right ]_d =  \left \{ z \in X : d(x,z)+d(z,y)=d(x,y)\right \}$
My question is, for metric spaces $(X, d)$ and $(X, d')$, what would be some sufficient conditions that the metrics $d$ and $d'$ need to satisfy such that it holds true that $\forall x,y \in X : \left [ x,y \right ]_d=\left [ x,y \right ]_{d'}$? That is, what would be some sufficient conditions that the metrics $d$ and $d'$ on $X$ need to satisfy such that for all points $x,y \in X$ the following holds true?
$d(x,p)+d(p,y)=d(x,y) \Leftrightarrow d'(x,p)+d'(p,y)=d'(x,y)$
A trivial such condition would be $\exists \lambda \in \mathbb{R}^+ : \forall x,y \in X : d'(x,y)=\lambda d(x,y)$. However, the $L^p$-metrics $d_p$ and $d_q$ on $\mathbb{R}^n$ satisfy that condition for all $p,q> 1$ and $n \in \mathbb{N}$ whilst not necessarily being a constant multiple of each other. It should be also noted that, for example, the metrics $d_1$ and $d_2$ on $\mathbb{R}^n$ do not satisfy that condition for all $n\geq 2$.
 A: Let $X$ be a set and let $d_1, d_2, ..., d_n$ be metrics on $X$. Let $\lambda_1, \lambda_2, ..., \lambda_n > 0$ and $\mu_1, \mu_2, ..., \mu_n > 0$ be real numbers. We will show then that the metrics $d : \left ( x,y \right ) \mapsto \sum_{i=1}^{n} \lambda _i d_i (x,y)$ and $d' : \left ( x,y \right ) \mapsto \sum_{i=1}^{n} \mu _i d_i (x,y)$ define the same metric segments.
Let $x, y \in X$ be points and let $p \in \left [ x,y \right ]_d$ be a point. By definition, it holds true that $d(x,p)+d(p,y)=d(x,y)$ which is equivalent to the statement $\sum_{i=1}^{n} \lambda _i \left [ d_i (x,p)+d_i (p,y)-d_i (x,y) \right ]=0$. However, by the assumption $\forall i \leq  n : \lambda_i > 0$ and by the triangle inequality, it follows that $\forall i \leq  n : \lambda _i \left [ d_i (x,p)+d_i (p,y)-d_i (x,y) \right ] \geq  0$. Therefore, it must hold true that $\forall i \leq  n : \lambda _i \left [ d_i (x,p)+d_i (p,y)-d_i (x,y) \right ]=0 $ which implies $\forall i \leq  n : d_i(x,p)+d_i (p,y)=d_i (x,y)$.
Having this in mind, it furthermore follows that $d'(x,p)+d'(p,y)=\sum_{i=1}^{n} \mu _i \left [ d_i (x,p)+d_i (p,y) \right ]=\sum_{i=1}^{n} \mu _i d_i(x,y)=d'(x,y)$ which is, by definition, equivalent to the statement $p \in \left [ x,y \right ]_{d'}$.
This establishes the relation $\left [ x,y \right ]_d \subseteq \left [ x,y \right ]_{d'}$. By symmetry, the relation $\left [ x,y \right ]_{d'} \subseteq \left [ x,y \right ]_d$ also follows which proves the statement $\left [ x,y \right ]_d=\left [ x,y \right ]_{d'}$.
