Let $d$ be a metric on a (say real) vector space $E$, with the property $$d(x,x+cy)=|c|d(x,x+y)$$ for all $x,y\in E$ and scalars $c$. I am trying to prove that $x\mapsto d(x,0)$ defines a norm.
The triangle inequality gives me trouble (the others are easy): I need $d(x+y,0)\leq d(x,0)+d(y,0)$ for all $x,y\in E$.
Some thoughts: setting $y=\lambda x$ and $c=-1$ in the 'property' gives $d(x,(1-\lambda)x)=d(x,(1+\lambda)x)$, in particular $d(x,0)=d(x,2x)$.
The usual triangle inequality for $d$ gives $d(x+y,0)\leq d(x+y,x)+d(x,0)$, so also $2d(x+y,0)\leq d(x+y,x)+d(x+y,y)+d(x,0)+d(y,0)$.
I need something like translation invariance, or express $d(x+y,0)$ in terms of $d(y,0)$ ('get rid of the sum').
Some hints/suggestions? Thanks!