The reverse triangle inequality

Let $d$ be metric on a set $X$ .
Show that $d$ satisfies the reverse triangle inequality:
$\forall x,y,z \in X , \, |d(x,z)-d(y,z)| \leqslant d(x,y).$

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Solution by darktoasterFR

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Soient $x,y,z \in X$

D’une part, par l’inégalité triangulaire : $d(x,z) \le d(x,y) + d(y,z)$
On a : $d(x,z) - d(y,z)\le d(x,y)$

D’une autre part, par l’inégalité triangulaire $d(y,z) \le d(y,x) + d(x,z)$
On a $d(y,z) - d(x,z) \le d(y,x)$
Par symétrie de $d$ : $d(y,z) - d(x,z) \le d(x,y)$

On a en conclusion : $|d(x,z) - d(y,z)| \le d(x,y)$

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