An inequality for metric spaces: $|d(x, z) − d(y, z)| \le d(x,y)$ Question : Prove $|d(x, z) − d(y, z)|$ is less than or equal to $d(x, y)$.


I know I have to use the triangle inequality but I'm just not sure how to apply it with a negative $d(y,x)$.
The claim is invariant under exchange of $x$ and $y$. Thus without loss of generality we can assume $d(x,z)\ge d(y,z)$. Then $|d(x,z)-d(y,z)|=d(x,z)-d(y,z)$, which is $\le d(x,y)$ by the triangle inequality.
First note that $|a-b|\leq |a|+|b|$\
Now, $|d(x,z)-d(y,z)|\leq |d(x,z)|+|d(y,z)|=d(x,z)+d(y,z)\leq d(x,y) (\because \text{triangle inequality}).$