Apologies in advance: I am new to math.stackexchange and do not know how to properly link.
It is well known that a norm $||\cdot||$ on a vector space $V$ induces a translation invariant, homogeneous metric $d$ on $V$ via $d(x,y) = ||x-y||$.
Conversely, it is easy to show that a translation invariant, homogeneous metric $d$ on a (real) vector space $V$ induces a norm via the map $v\to d(v,0)$.
In their reply to Not every metric is induced from a norm , the user C-Star-W-Star [user:79762] states:
"Every homogeneous metric induces a norm via: \begin{equation}||x|| := d(x,0). \end{equation}
C-Star-W-Star then shows that the metric induced by this norm is the original metric only if the original metric is translation invariant.
My question: How can one show the triangle inequality holds for this $||x||$ without assuming that $d$ is translation invariant?
