Prove that $d$ is not a metric on an open subset $U$ of $\mathbb{R}^n$ I've got the map $d: \mathbb{R}^n \to \mathbb{R}^n$ defined as follows:
$$
d(x,y)=\sum\limits_{i=1}^n \frac{|x_i-y_i|}{|x_i|+|y_i|}
$$
Restrict $d$ to the subset $U=\{x \in \mathbb{R}^n: \, x_i \neq 0 \, \forall i=1,\dots,n\}$. I'm asking whether such a map is a metric on $U$. The problem is clearly given by the triangle inequality. Some time ago I read in a book (I don't remember the title) that it is not a metric. However, I'm not able to find a simple counterexample. Can you help me, please
I tried lots of cases, but I observe that triangle inequality always holds. Let us reason for only a variable: 
$$
d(x,y)=\frac{|x-y|}{|x|+|y|}
$$
The previous becomes
$$
d(x,y)=\frac{|1-\frac{y}{x}|}{1+|\frac{y}{x}|}.
$$
Let $t=\frac{y}{x}$. I may study the monotonicity of the function $\frac{|1-t|}{1+|t|}$. This is not an increasing function, so triangle inequality cannot hold. Is my argument correct?
 A: I'll show it is a metric on $\mathbb R$.
Proposition: $d(x,y):=\frac{\vert x-y\vert}{\vert x\vert+\vert y\vert}$
is a metric. I'll prove the triangle inequality.
Notice this metric image is $[0,1]$.
We need to show: $\forall x,y,z\in\mathbb{R}\ \ d(x,z)\leq d(x,y)+d(y,z)$
Assume WLOG $\vert x\vert  \geq\vert z\vert$.
If $z=0$:
$d(x,0)=1=d(y,0)\leq d(x,y)+d(y,z)$
Therefore, from now on, I'll prove the claim for $z\neq0$.
We can divide $x,y,z$ by $z$. Therefore it's sufficient to prove:
$\forall x',y'\in\mathbb{R}\ \ d(x',1)\leq d(x',y')+d(y',1)$= $\frac{\vert x'-1\vert}{\vert x'\vert+\vert1\vert}\leq\frac{\vert x'-y'\vert}{\vert x'\vert+\vert y'\vert}+\frac{\vert y'-1\vert}{\vert y'\vert+\vert1\vert}$
Case 1: $\vert z\vert=1\geq\vert y'\vert$.
$$ \begin{align}
d(x',z) &=\frac{\vert x'-1\vert}{\vert x'\vert+\vert1\vert}\\
&\leq\frac{\vert x'-1\vert}{\vert x'\vert+\vert y'\vert}\\
&\leq\frac{\vert x'-y'+y'-1\vert}{\vert x'\vert+\vert y'\vert}\\
&\leq\frac{\vert x'-y'\vert+\vert y'-1\vert}{\vert x'\vert+\vert y'\vert}\\
&=\frac{\vert x'-y'\vert}{\vert x'\vert+\vert y'\vert}+\frac{\vert y'-1\vert}{\vert x'\vert+\vert y'\vert}\\
&\leq\frac{\vert x'-y'\vert}{\vert x'\vert+\vert y'\vert}+\frac{\vert y'-1\vert}{\vert1\vert+\vert y'\vert}\\
&=d(x',y')+d(y',1)\\
\end{align}$$
We can now assume $\vert y'\vert\geq\vert z\vert$.
Case 2: $x'<0$
$y'\geq0\implies d(x',y')=1\implies d(x',1)\leq1\leq d(x',y')+d(y',1)$
$y'<0\implies d(y',1)=1\implies d(x',1)\leq1\leq d(x',y')+d(y',1)$
Case 3: $x'\geq0\implies x'\geq1(x'\geq z=1)$
If $y' \leq 0$, then by case 2, $d(y',1) = 1$. since $d(x',1)\leq
1$ we are done
Therefore, $x',y'$ are positive. We are left with two cases. $y'\geq x',y'<x'$. 
Because $x',y'$ are positive and greater than 1: $d(x',1)=\frac{x'-1}{x'+1},d(y',1)=\frac{y'-1}{y'+1}.$
Define a function $f:[1,\infty)\rightarrow[0,1]$, by $f(x'):=\frac{x'-1}{x'+1}$.
$f$ monotonicly increasing (on it's range) because it's derivative $f'(x)=\frac{2}{{(x-1)}^2}$ is positive (Well defined on the function range). Also note that $f(x')=d(x',1)$ on $[1,\infty)$.
The case $y'\geq x'$ is immediate, since $f(x')$ is monotonicly increasing:
$d(x',1)=f(x')\leq f(y')\leq f(y')+d(x',y')\leq d(y',1)+d(x',y')$
Finally we are left with a final case: $x'\geq y'\geq1$. 
Claim: $f(y'+t)-f(y')=\frac{2t}{y'(y'+t+2)+t+1}$
Proof:
$$\begin{align}
& f(y'+t)-f(y')=\frac{y'+t-1}{y'+t+1}-\frac{y'-1}{y'+1}\\
&=\frac{(y'+t-1)(y'+1)}{(y'+t+1)(y'+1)}-\frac{(y'-1)(y'+t+1)}{(y'+1)(y'+t+1)}
\end{align}$$ 
Only the numerator: 
$$\begin{align}
&(y'+t-1)(y'+1)-(y'-1)(y'+t+1)\\
& =y'^{2}+ty'-y'+y'+t-1-(y'^{2}+ty'+y'-y'-t-1)=2t 
\end{align}$$
Only the denominator:
$$\begin{align}
&(y'+t+1)(y'+1)=y'^{2}+y'+ty'+1+y'+t=y'(y'+t+2)+t+1\ \ \square
\end{align}$$
Claim 2: $f(y'+t)-f(y')\leq d(x',y')$
I'll show
$$
\begin{align*}
&f(y'+t)-f(y') \leq d(x',y')\\
&\iff\frac{2t}{y'(y'+t+2)+t+1}\leq\frac{t}{y'+y'+t}\\
&\iff\frac{2}{y'(y'+t+2)+t+1}\leq\frac{1}{y'+y'+t}\\
&\iff 2(2y'+t)\leq y'(y'+t+2)+t+1\\
&\iff2y'+t\leq y'(y'+t)+1.\\
\end{align*}$$
$t\leq y't$ since $y'\geq 1$.
$2y'\leq y'^{2}+1\iff0\leq y'^{2}-2y'+1\iff0\leq(y'-1)^{2}\iff $always!
Therefore this case always holds!
Conclusion: d is a satisfies the triangle inequality.
This paper proves the "Normalized Symmetric Difference" is a metric, Which is generalization of this case, to sets. 
Sorry about the mistakes on the previous post. This question drove me crazy!
May I ask where it is taken from?
