triangle inequality for a certain norm 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!
 A: Be it homework or not – this is a very fine problem indeed!
It is enough to consider a plane $E$ with points $z=(x,y)$. Let $|z|:=\sqrt{x^2+y^2}$ denote the euclidean norm.
(i) Claim: For each line $g\subset E$ there is a constant $\lambda_g$ such that for all $z$, $w\in g$ one has
$$d(z,w)=\lambda_g\ |z-w|\ .$$
Proof. Let
$$g:\quad t\mapsto z(t)=a + t\ u\ ,\qquad |u|=1$$
be a parametrization of $g$ with euclidean arc length as parameter and put
$$d^*(s,t):=d\bigl(z(s),z(t)\bigr)\ .$$
Then by the basic property of $d(\cdot,\cdot)$ we have for arbitrary $s$, $t\in{\mathbb R}$:
$$d^*(s,t)=d^*\bigl(s,s+(t-s)\bigr)=|t-s|d^*(s,s+1)$$
and similarly
$$d^*(s,t)=d^*(t,s)=d^*\bigl(t,t+(s-t)\bigr)=|t-s|d^*(t,t+1)\ .$$
For $t:=0$ the last two equations imply $|s|d^*(s,s+1)=d^*(s,0)=|s|d^*(0,1)$. It follows that $d^*(s,s+1)=d^*(0,1)=:\lambda_g$ for arbitrary $s\in{\mathbb R}$, whence
$$d\bigl(z(s),z(t)\bigr)=d^*(s,t)=\lambda_g |t-s|=\lambda_g\bigl|z(t)-z(s)\bigr|\ ,$$
as claimed.
(ii) Claim: The scaling factor $\lambda_g$ depends only on the direction vector $u=(\cos\phi,\sin\phi)$ of $g$. In fact there is a continuous function $\lambda:\ S^1\to{\mathbb R}_{>0}$ such that $\lambda_g= \lambda(\phi)$.
Proof. It is enough to consider the following situation: Let $g_1$ be the line $y=1$ and $g_2$ be the line $y=-1$, and let $p_i:=\lambda_{g_i}$. We have to show that $p_1=p_2$. Consider for small $\phi >0$ the line
$$ g_\phi:\quad t \mapsto t \ (\cos\phi,\sin\phi)\ .$$
The line $g_\phi$ has a certain scaling factor $\lambda(\phi)$ and intersects $g_1$ at the point $b:=(\cot\phi, 1)$. Put $(0,0)=:o$, $(0,1)=:a$. Then we have by the triangle inequality for $d$:
$$d(a,b)-d(o,a)\leq d(o,b)\leq d(o,a)+d(a,b)$$
or
$$p_1\cot\phi - d(o,a)\leq \lambda(\phi)\ {1\over\sin\phi}\leq d(o,a)+p_1\cot\phi\ .$$
Multiplying by $\sin\phi>0$ we obtain
$$p_1 \cos\phi -d(o,a)\sin\phi \leq\lambda(\phi)\leq p_1\cos\phi+d(o,a)\sin\phi\ ,$$
from which we conclude $\lim_{\phi\to 0+}\lambda(\phi)=p_1$. By symmetry (arguing on the left side) we also have $\lim_{\phi\to 0-}\lambda(\phi)=p_1$, so that in fact $\lim_{\phi\to0}\lambda(\phi)=p_1$. Using symmetry again it follows that $p_2=p_1$. Since the lines $g_1$, $g_2$ had nothing special about them it follows that any two parallel lines have the same scaling factor $\ \bigl(=:\lambda(\phi)\bigr)$; in particular the scaling factor of $g_0\colon\ y=0$ is also $\ =p_1$, which proves the continuity of the function $\lambda(\cdot)$. In particular, this function is bounded away from $0$ and $\infty$.
(iii) It follows from (i) and (ii) that the metric $d(\cdot,\cdot)$ is translation invariant. We now define the associated norm by
$$\|z\|\ :=\ d(o,z)\ .$$
The identity $\|\alpha\ z\|=|\alpha|\ \|z\|$ is obvious, and
$$\|z+w\|=d(0,z+w)\leq d(o,z)+d(z,z+w)=d(o,z)+d(o,w)=\|z\|+\|w\|$$
proves the triangle inequality for the norm.
