Conjecture: the function $d(x, y):=\frac{||x-y||}{\max(||x||, ||y||)}$ is a distance I make the following conjecture: the function 
$$
d(x, y):=\frac{||x-y||}{\max(||x||, ||y||)}
$$
is a distance on $H$, where $H$ is a normed vector space or a Hilbert space, and $x, y \in H$ (the function $d$ is defined to be $0$ in the case $x=y=0$). Note that $d$ is scale invariant, i.e., $d(\lambda x, \lambda y)=d(x, y)$ for $0 \neq \lambda \in \mathbb{R}$. The property of $d$ which needs to be explicitly proved or disproved is the triangle inequality
$$
d(x, y) \leq d(x,z)+d(z,y).
$$
The triangle inequality (TI) can be easily proved for $H=\mathbb{R}$; moreover, due to scale invariance, it is sufficient to prove it for $||x||,||y||, ||z|| \leq 1$. The TI has been numerically tested by a program which has generated $10^8$ triples of random points uniformly distributed in $[-1, 1]^3$, and the same number in $[-1,1]^6$: all the generated triples satisfied the TI. This test supports therefore the conjecture for  $H=\mathbb{R}^3$ and $H=\mathbb{R}^6$. Since the subspace generated by three linearly independent vectors of a real (complex) Hilbert space is isometrically isomorphic to $\mathbb{R}^3$ ($\mathbb{R}^6$), the numerical test supports the conjecture also for a generic Hilbert space*. 
Does somebody know if this conjecture has been already proved, or is able to prove (or disprove) it?


*

*Before posting this question, I exchanged some e-mail with prof. Egor Makimenko, of the Instituto Politécnico Nacional, México. I did by myself a program for the numerical test of the TI, but the test cited above has been performed by a program that prof. Maximenko sent me. Moreover, the generalization from $\mathbb{R}^3$-$\mathbb{R}^6$ to a generic Hilbert space is due to prof. Maximenko.

 A: Let the norm on the plane be the taxicab norm, norm of $(x,y)$ being $|x|+|y|.$ This induces a metric (satisfying triangle inequality etc.) on the metric space $\mathbb{R}^2$, making it a normed linear space.
But with your metric obtained by division of max norms, take the points $x,y,z$ as $(1,2),(2,1),(2,2)$ respectively. Then the usual distances (before division) are $2$ for $x,y$ and each usual distance for both $x,z$ and $y,z$ is $1.$ The norms of $x,y,z$ are respectively $3,3,4$, so that after division by max norms we get
$$d(x,y)=2/3, \ d(x,z)=d(z,y)=1/4.$$
This is against the triangle inequality since $2/3>1/4+1/4.$
So in conclusion it looks to me like the "metric" you propose does not in general satisfy the triangle inequality on an arbitrary normed linear space. I would still like to see if it holds in Euclidean space, since some features of taxicab metric are peculiar.
A: We know that $p(x,y):= \|x-y\|$ is a metric. Hence we get for free
$$\|x-z\|\leq \|x-y\|+\|y-z\|.$$
Remark: The case where exactly one of the vectors is the zero vector is trivial because we get an inequality of the form:
$$1\leq 1+\varepsilon.$$
If exactly two are zero, we get either:
$$1\leq 1+0,$$
OR
$$0\leq 1+1.$$
If all three are zero:
$$0\leq 0+0.$$
Now consider the case $0<\|x\|\leq \|y\| \leq \|z\|$:
\begin{align}
\max(\|x\|,\|z\|)&=\|z\|, \\
\max(\|x\|,\|y\|)&=\|y\|, \\
\max(\|y\|,\|z\|)&=\|z\|.
\end{align}
So, dividing through the first inequality by $\max(\|x\|,\|z\|)$ we get
$$\frac{\|x-z\|}{\max(\|x\|,\|z\|)}\leq \frac{\|x-y\|}{\max(\|x\|,\|z\|)} + \frac{\|y-z\|}{\max(\|x\|,\|z\|)},$$
but $$d(x,y)\geq \frac{\|x-y\|}{\max(\|x\|,\|z\|)} \text{  and  } d(y,z)=\frac{\|y-z\|}{\max(\|x\|,\|z\|)}.$$
Hence, the triangle inequality is satisfied.
I am optimistic that the other 5 cases can be proven similarly. Don’t hold me to it! I hope you can make some progress from here.
A: The inequality purposed by BGA fails also in $(\mathbb{R}^2,\|\cdot\|_\infty)$,
but it seems to be true for the norms induced by inner products.
Hypothesis I: If $X$ is an inner product space and
$\|\cdot\|$ is the norm induced by the inner product,
then for every $x,y,z\in X$
$$\|x-y\|\,\|z\| \le \|x-z\|\,\|y\| + \|z-y\|\,\|x\|.$$
I don't know how to prove this hypothesis in general situation,
but it can be easily proved in $X=\mathbb{R}^1$ and it passes some
numerical tests in $X=\mathbb{R}^3$.
The difficult case ($\|z\|\ge\|x\|\ge\|y\|$) of the discussed triangular inequality $d(x,y)\le d(x,z)+d(z,y)$ follows from Hypothesis I.
