norm induced by distance implies translation invariant distance? In his book linear Analysis , Bollobás says 

Given a metric $d$ on a vector space $X$, setting $||x||=d(x,0)$ defines a norm on $X$ iff $d(x,y)=d(x+z,y+z)$ and $d(\lambda x,\lambda y)=|\lambda|d(x,y)$ for all $x,y,z \in X$ and scalar $\lambda$.

The part where the distance is homogeneous and translation invariant implies that $||x||=d(x,0)$ is a norm is quite easy and I'm ready with that.
I'm stuck in the other side. In particular, I have no idea how to prove that the distance must be translation invariant (homogeneity is simple).
Any advice would be a great help.
 A: The statement is not true.  
Lets consider $\mathbb{R}^2$ as a metric space using the so-called "French railway distance" $\delta$. The distance $\delta(A,B)$ from a point $A$ to $B$ is the usual (euclidean) distance $\overline{AB}$ if they lie on the same ray from the origin $O$. Otherwise, $\delta(A,B)=\overline{AO}+\overline{OB}$.
We can write
$\delta(A,B)=\begin{cases}
|A|+|B|,  & \text{if $A\neq \lambda B$} \\
|A-B|, & \text{if $A= \lambda B$ for some $\lambda$}
\end{cases}$
I will show that $\delta$ is a distance function. In fact:


*

*$\delta(A,B) \geq 0\ \forall\ A,B\in \mathbb{R}^2$ and $\delta(C,C)=0\ \forall C \in \mathbb{R}^2$  

*$\delta(A,B)=\delta(B,A)$. This is clear for any case.

*$\delta(A,B)\leq \delta(A,C)+\delta(C,B)$. Here we have a lot of cases:


*

*$\fbox{$A\neq \lambda B$. $C\neq \lambda A$ and $C\neq \lambda B.$}$ Then $\delta(A,B)=|A|+|B|\leq |A|+|C|+|C|+|B|=\delta(A,C)+\delta(C,B)$. 

*$\fbox{$A\neq \lambda B$, $C\neq \lambda A$ but $C= \lambda B$ for some $\lambda$.}$ Then $\delta(A,B)=|A|+|B|=|A|+|B-C+C|\leq |A|+|C|+|B-C|=\delta(A,C)+\delta(C,B)$.  

*$\fbox{$A\neq \lambda B$, $C\neq \lambda B$ but $C= \lambda A$ for some $\lambda$.}$ As in the previous case, here inequality is also true by symmetry. 

*$\fbox{$A\neq \lambda B$, $C= \lambda_1 A$ for some $\lambda_1$ and $C= \lambda_2 B$ for some $\lambda_2$.}$ Cannot be possible.  

*$\fbox{$A= \lambda B$ for some $\lambda$, $C\neq \lambda_1 A$ (and then $C\neq \lambda_2 B$).}$ Then $\delta(A,B)=|A-B|\leq |A|+|B|\leq |A|+|C|+|B|+|C|=\delta(A,C)+\delta(C,B)$.  

*$\fbox{$A= \lambda B$ for some $\lambda$, $C= \lambda_1 A$ for some $\lambda_1$ (and then $C= \lambda_2 B$ for some $\lambda_2$).}$ Then $\delta(A,B)=|A-B|=|A-C+C-B|\leq |A-C|+|C-B|=\delta(A,C)+\delta(C,B)$.



And it is homogeneous, because clearly $\delta(\rho A,\rho B)=|\rho|\delta(A,B)$ for any case.
However, it is not translation invariant. If we take A=(3,0), B=(0,3), then $\delta(A,B)=6$ and translating by $C=(4,0)$ we have $A+C=(7,0)$ and $B+C=(4,3)$ so $\delta(A+C,B+C)=7+5=12.$
Now, we see that setting $||A||=d(A,0)$ defines a norm on $\mathbb{R}^2$. 
In fact:  
i)$||A||=0$ iff $\delta(A,0)=0$ iff $A=0$.
ii)$||\lambda A||=\delta(\lambda A,0)=|\lambda A| = |\lambda||A| = |\lambda|\delta(A,0) =|\lambda|||A||$.
iii)$||A+B||=\delta(A+B,0)=|A+B|\leq |A|+|B|= \delta(A,0)+\delta(B,0)=||A||+||B||$.
