I have to determine which of the following define a metric on $\Bbb R \,\,$? I am stuck on the following problem:  

Determine which of the following define a metric on $\Bbb R$:
  
  
*
  
*$d(x,y)=\frac{|x-y|}{1+|x-y|}$   
  
*$d(x,y)=|x-2y|+|2y-x|$  
  
*$d(x,y)=|x^2-y^2|$     

MY ATTEMPT: 
In each of the aforementioned cases, $d(x,y) \ge 0 $ and $d(x,y)=d(y,x).$ So, I have to check the triangle inequality.  
For option 1, $$d(x,y)=\frac{|x-y|}{1+|x-y|} \implies d(x,y) \le \frac {|x-y|}{|x-y|}=1$$ and hence $d(x,y) =1 \le d(x,z)+d(z,y)=2.$ So, option 1 defines a metric on $\Bbb R$.  
For option 2, I can not prove triangle inequality and I need help here.   
For option 3, we see that
$$\begin{align*}
d(x,y)&=|x^2-y^2|\\\\
& =|(x^2-z^2)+(z^2-y^2)| \\\\
&\le |x^2-z^2|+|z^2-y^2|\\\\
&=d(x,z)+d(z,y)
\end{align*}$$ and so option 3 defines a metric on $\Bbb R$.  
Am I right? Thanks in advance for your time. 
 A: For 2) $d(1,2)=6$ while $d(2,1)=0$, so it fails to be a metric on two accounts of the definition. 
For 1) your proof is incorrect as clearly $d(x,y)$ does not have to be equal to $1$ for all $x,y$ (in fact, it's never equal to $1$). You need to show that $d(x,y)\le d(x,z)+d(z,y)$, so knowing that $d(u,v)\le 1$ does not help at all. Try proving it in general: for any metric $d$ whatsoever, the function $\rho(x,y)=\frac{d(x,y)}{1+d(x,y)}$ is a metric. 
For 3) your proof of the triangle inequality is correct but $d(1,-1)=0$, thus it is not a metric space. 
A: This is for the 1)
d(x,y)=$\frac{|x-y|}{1+|x-y|}$
You need to prove the triangular inequality, i.e.
$\frac{|x-y|}{1+|x-y|}\le$$\frac{|x-z|}{1+|x-z|}$+$\frac{|z-y|}{1+|z-y|}$
we have $1+|x-y|\leq 1+|x-z|+|z-y|$
$1-\frac{1}{1+|x-y|}\leq$$1-\frac{1}{1+|x-z|+|z-y|}$
$\frac{|x-y|}{1+|x-y|}\leq$$\frac{|x-z|+|z-y|}{1+|x-z|+|z-y|}$
$\leq\frac{|x-z|}{1+|x-z|+|z-y|}$+$\frac{|z-y|}{1+|x-z|+|z-y|}$
$\leq\frac{|x-z|}{1+|x-z|}$+$\frac{|z-y|}{1+|z-y|}$
This way you show that 1) satisfies the triangular inequality apart from the other two conditions to be a metric.
2) is not a metric, because |x-2y|$\ne$|y-2x| always, look up why. 
