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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{)}} $

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    What have you tried so far? Did you use the axioms for metrics? In that case which axioms have you tried to use? – 2017-02-16
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    I jusst tried with the third axiom and the absolute value properties and I couldn't reach – 2017-02-16
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    Which is the third? Different sources number them differently. Wikipedia numbers them (1) non-negativity, (2) identity of indiscernibles, (3) symmetry, (4) sub-additivity (triangle inequality). You certainly need to use the latest. – 2017-02-16
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    Yes the fourth one – 2017-02-16

1 Answers 1

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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).