Metric spaces and the absolute value I tried with the absolute value properties to solve it but I couldn't find it.
Let  $
{\mathrm{(}}{X}{\mathrm{,}}\mathit{\rho}{\mathrm{)}}
$
be a metric space :
a) for $
\mathrm{\forall}\hspace{0.33em}{x}{\mathrm{,}}{y}{\mathrm{,}}{z}
$  show that :$
\left|{\mathit{\rho}{\mathrm{(}}{x}{\mathrm{,}}{z}{\mathrm{)}}\mathrm{{-}}\mathit{\rho}{\mathrm{(}}{y}{\mathrm{,}}{z}{\mathrm{)}}}\right|\mathrm{\leq}\mathit{\rho}{\mathrm{(}}{x}{\mathrm{,}}{y}{\mathrm{)}}
$
b)for $
\mathrm{\forall}\hspace{0.33em}{x}{\mathrm{,}}{y}{\mathrm{,}}{z}{\mathrm{,}}{w}
$ show that: $
\left|{\mathit{\rho}{\mathrm{(}}{x}{\mathrm{,}}{y}{\mathrm{)}}\mathrm{{-}}\mathit{\rho}{\mathrm{(}}{z}{\mathrm{,}}{w}{\mathrm{)}}}\right|\mathrm{\leq}\mathit{\rho}{\mathrm{(}}{x}{\mathrm{,}}{z}{\mathrm{)}}\mathrm{{+}}\mathit{\rho}{\mathrm{(}}{y}{\mathrm{,}}{w}{\mathrm{)}}
$
 A: You could rewrite the inequalities and drop the absolute values. For example the first inequality is the same as:
$$-\rho(x,y) \le \rho(x,z) - \rho(y,z) \le \rho(x,y)$$
Then you can split that in two and shuffle around the terms.
To elaborate: by the triangle inequality we have:
$$\rho(x,z) \le \rho(x,y) + \rho(y,z)$$
$$\rho(x,z) - \rho(y,z) \le \rho(x,y)\tag1$$
also by the triangle inequality we have:
$$\rho(y,z) \le \rho(y,x) + \rho(x,z) = \rho(x,y) + \rho(x,z)$$
$$\rho(y,z)-\rho(x,y) \le \rho(x,z)$$
$$-\rho(x,y) \le \rho(x,z)-\rho(y,z)\tag2$$
Now combining (1) and (2) we get:
$$-\rho(x,y) \le \rho(x,z)-\rho(y,z) \le \rho(x,y)$$
Now since $-a \le b \le a$ is eqivalent to $|b|\le a$ we have:
$$|\rho(x,z)-\rho(y,z)| \le \rho(x,y)$$
The second inequality is proven in similar manner, but you need to extend the triangle inequality to involve more intermediate points:
$$\rho(a,d) \le \rho(a,b) + \rho(b,c) + \rho(c,d)$$
(you can actually generalize this to arbitrary number of intermediate points).
