If $d$ is a metric, is $d(x,y)/(1+d(x,0)+d(y,0))$ a metric?

I now that one can show that if $d$ is a metric on a vectorspace $X$ then so is $$\varrho(x,y):=\frac{d(x,y)}{1+d(x,y)}.$$ This easily follows from the fact that the function $s \mapsto \frac{s}{1+s}$ is increasing on $\mathbb{R}$.

But does the same hold true for $$\nu(x,y):=\frac{d(x,y)}{1+d(x,0)+d(y,0)},$$ i.e. is this a metric, in the case $X=\mathbb{R}$ and $d(x,y)=|x-y|$?

Of course one could use the reasoning above to see that $$\frac{d(x,y)}{1+d(x,0)+d(y,0)} \leq \frac{d(x,y)}{1+d(x,y)} \leq \frac{d(x,z)+d(z,y)}{1+d(x,z)+d(z,y)} = \frac{d(x,z)}{1+d(x,z)+d(z,y)}+\frac{d(z,y)}{1+d(x,z)+d(z,y)}$$ but that leads nowhere since the quantity $d(x,z)+d(z,y)$ tells us nothing about the quantity $d(x,0)+d(z,0)$.

Can anybody give me a hint on how to do that? Thank you very much in advance!

We are trying to prove that for all $x,y,z\in\mathbb R$ the inequality $$\frac{|x-y|}{1+|x|+|y|}\leq\frac{|x-z|}{1+|x|+|z|}+\frac{|z-y|}{1+|z|+|y|}\tag{1}$$ holds. Let $x$ and $y$ be fixed. We may assume without loss of generality that $x<y$ (interchanging $x$ and $y$ does not change the inequality and the $x=y$ case is trivial). Let $a\leq b\leq c$ be the numbers $0,x,y$, ordered by size. Then, the function $f:\mathbb R\to\mathbb R$ defined by $$\begin{align}f(z)&=\left(\frac{|x-z|}{1+|x|+|z|}+\frac{|z-y|}{1+|z|+|y|}-\frac{|x-y|}{1+|x|+|y|}\right)(1+|x|+|z|)(1+|z|+|y|)\\&=|z-x|(1+|y|+|z|)+|z-y|(1+|x|+|z|)-\frac{|x-y|}{1+|x|+|y|}(1+|x|+|z|)(1+|z|+|y|)\end{align}$$ is quadratic on each of the (possibly degenerate) intervals $(-\infty,a],[a,b],[b,c],[c,\infty)$. Proving the inequality $(1)$ for a particular $z\in\mathbb R$ is equivalent to proving that $f(z)\geq 0$ for that $z$.

Note that $f(x)=f(y)=0,$ so it is sufficient to prove that these are the global minima of $f$. Note that the RHS of $(1)$ goes to $2$ as $z\to\pm\infty$, while the LHS remains less than $1$ by the triangle inequality, so the inequality is true for all $z$ that are large enough in absolute value. Therefore, it remains to verify that the local minima of $f$ are non-negative (if any exist) and that $f(0)\geq 0$ (this is the only remaining singular point).

To verify $f(0)\geq 0$, simply verify the inequality for $z=0$: $$\frac{|x-y|}{1+|x|+|y|}\leq\frac{|x|+|y|}{1+|x|+|y|}=\frac{|x|}{1+|x|+|y|}+\frac{|y|}{1+|x|+|y|}\leq\frac{|x|}{1+|x|}+\frac{|y|}{1+|y|}.$$

Next, the function $f$ is differentiable everywhere except possibly at $0,x,y$. Recall that the derivative of the absolute value function $z\mapsto|z|$ is the sign function $z\mapsto\operatorname{sgn}(z)$, where it exists. Therefore we may compute the derivative of $f$: $$f'(z)=\operatorname{sgn}(z-x)(1+|y|+|z|)+|z-x|\operatorname{sgn}(z)+\operatorname{sgn}(z-y)(1+|x|+|z|)+|z-y|\operatorname{sgn}(z)-\frac{|x-y|}{1+|x|+|y|}\operatorname{sgn}(z)(1+|z|+|y|)-\frac{|x-y|}{1+|x|+|y|}(1+|x|+|z|)\operatorname{sgn}(z)$$ and the second derivative is as follows: $$\begin{align}f''(z)&=\operatorname{sgn}(z-x)\operatorname{sgn}(z)+\operatorname{sgn}(z-x)\operatorname{sgn}(z)+\operatorname{sgn}(z-y)\operatorname{sgn}(z)+\operatorname{sgn}(z-y)\operatorname{sgn}(z)-\frac{|x-y|}{1+|x|+|y|}\operatorname{sgn}(z)\operatorname{sgn}(z)-\frac{|x-y|}{1+|x|+|y|}\operatorname{sgn}(z)\operatorname{sgn}(z)\\&=2\operatorname{sgn}(z)(\operatorname{sgn}(z-x)+\operatorname{sgn}(z-y))-2\frac{|x-y|}{1+|x|+|y|}.\end{align}$$ A quadratic function can achieve a minimum only at a point where the second derivative is non-negative. The last calculation tells us that this can only happen if $$\operatorname{sgn}(z)(\operatorname{sgn}(z-x)+\operatorname{sgn}(z-y))>0.$$ This is true precisely if $z<x$ and $z<0$ or $z>y$ and $z>0$.

These two cases are precisely the two cases when the values of the $\operatorname{sgn}$ function that appear in our calculation of $f'$ are the same. We claim that $f'(z)\neq 0$ for these $z$ (implying that no other local minima can occur). Since the signs are all the same, this reduces to showing that $$(1+|y|+|z|)+|z-x|+(1+|x|+|z|)+|z-y|\neq\\\neq\frac{|x-y|}{1+|x|+|y|}(1+|z|+|y|)+\frac{|x-y|}{1+|x|+|y|}(1+|x|+|z|).$$ But we have already noticed in the beginning that $$\frac{|x-y|}{1+|x|+|y|}<1.$$ This implies that $$(1+|y|+|z|)+|z-x|+(1+|x|+|z|)+|z-y|>(1+|y|+|z|)+(1+|x|+|z|)$$ and $$\frac{|x-y|}{1+|x|+|y|}(1+|z|+|y|)+\frac{|x-y|}{1+|x|+|y|}(1+|x|+|z|)<(1+|y|+|z|)+(1+|x|+|z|)$$ and the inequality is established.