Assume we have a vector space, $S$, with a distance function $D: S\times S \rightarrow \mathbb{R}^+$ with the following four assumptions about $D$:
(1) $D(x,y) = 0$ if and only if $x=y$
(2) $D(x,y) = D(y,x)$
(3) $D(x,y) \leq D(x,z) + D(z,y)$
(4) $D(\lambda x, \lambda y) = \lambda D(x,y)$ for $\lambda \in \mathbb{R}^+$.
where (4) makes our distance function $D$ into a Minkowski distance. Here is the (intuitively very obvious) thing which I can't prove:
If $x,y \in S$ and $z \equiv \lambda x + (1-\lambda) y$, with $\lambda \in [0,1]$, then $D(x,y) = D(x,z) + D(z,y) $.
I want to prove this for a general distance function assuming these axioms. I can't quite show how, despite this just being a simple statement about points on a line.
This is not true. For instance, define $D$ on $\mathbb{R}$ by $D(x,x)=0$ and $D(x,y)=|x|+|y|$ for $x\neq y$. This satisfies all your axioms but $D(1,3)\neq D(1,2)+D(2,3)$.
It is true if you additionally assume that $D$ is translation-invariant (i.e., $D(x+y,x+z)=D(y,z)$). In that case you can translate by $-x$ to assume $x=0$ and so it sufficies to prove that $D(0,y)=D(0,ty)+D(ty,y)$ for $t\in[0,1]$. For this, note that $D(0,ty)=tD(0,y)$ and $D(ty,y)=D(0,(1-t)y)=(1-t)D(0,y)$.