Brute force way to show that $\rho(x,y) = \min\{1, d(x,y)\}$ is a metric Following a hint in Short proof that $\rho^\prime(x,y) = \min\{1,\rho(x,y)\}$ is a metric 
I would like to use the brute force method to show that the standard bounded metric is a metric
$$\rho(x,y) = \min\{1, d(x,y)\}$$
By brute force I mean to substitute the definition of $\min$ operator which is:
$$\min(a,b) = \dfrac{a+b - |a-b|}{2}$$
So let $x,y,z \in (X, \rho)$
\begin{align}
\rho(x,z) &= \min\{1, d(x,z)\}\\
          &= \dfrac{1+d(x,z) - |1-d(x,z)|}{2}\\
          & = \dfrac{1}{2} + \dfrac{d(x,z)}{2} - \dfrac{|1-d(x,z)|}{2}\\
          & \leq \dfrac{1}{2} + \dfrac{d(x,y)}{2} + \dfrac{1}{2} + \dfrac{d(y,z)}{2}  - \dfrac{|1-d(x,z)|}{2} \tag{obvious}
\end{align}
If I can show that 

$$- \dfrac{|1-d(x,z)|}{2} \leq - \dfrac{|1-d(x,y)|}{2} + -
 \dfrac{|1-d(y,z)|}{2}$$

Then I am done. 
However, this seems to be really messy and difficult to show. How can I show that the above inequality holds?
 A: you have to check three things.
We have $\rho(x,y)=\rho(y,x)$ trivially (since $\delta(x,y)=\delta(y,x)$)
We also have $\rho(x.y)=\min(1,\delta(x,y))=0\iff\delta(x,y)=0\iff x=y$
The only tricky part is the triangle inequality.
We have to show $\rho(x,y)+\rho(y,z)\leq \rho(x,z)$.
if $\rho(x,y)\neq \delta(y,z)$ or $\rho(y,z)\neq \delta(y,z)$ we have $\rho(x,y)+\rho(x,z)\geq 1 \geq \rho(x,z)$ Otherwise:
$\rho(x,y)+\rho(y,z)=\delta(x,y)+\delta(y,z)\geq\delta(x,z)\geq\min(1,\delta(x,z))=\rho(x,z)$
A: In this case worth to calculate step by step. We have eight (8) possibilites.
Look: we want to show that $\rho^\prime(x,z) \leq \rho^\prime(x,y) + \rho^\prime(y,z)$, ok?! 
This is that $\min\{1, d(x,z)\} \leq \min\{1, d(x,y)\} + \min\{1, d(y,z)\}$. 


*

*Notice that, if $\rho^\prime(x,z)=d(x,z)=\min\{1, d(x,z)\}$ (this hypothesis say that $d(x,z)\leq 1$ it is very important for you understand the solution following) than we have four (4) possibilites:


*

*$d(x,z) \leq 1+ 1$;

*$d(x,z) \leq 1+ d(y,z);$

*$d(x,z) \leq d(x,y) +1;$

*$d(x,z) \leq d(x,y) + d(y,z);$ (it's true by hypothesis that $d$ is a metric)


*Now, if $\rho^\prime(x,z)=1=\min\{1, d(x,z)\}$ we have:


*

*$1 \leq 1+ 1;$

*$1 \leq d(x,y) + 1;$

*$1 \leq 1+ d(y,z);$

*$1 \leq d(x,y)+ d(y,z);$ this is true by hipothesis $\rho^\prime(x,z)=1\leq d(x,z)\leq d(x,y)+d(y,z);$



However, by 1-8 we proof triangle inequality. 
