Proof for absolute value inequality of three variables: $|x-z| \leq |x-y|+|y-z|$

Question

$|x-z| \leq |x-y|+|y-z|$

We know that both LHS and RHS are non negatives. So, I thought of proving this by comparing the squares of both sides but can't advance beyond that step. Any help would be appreciated.

Answer 1

If two of the variables are equal, it is easy. If they are not, there are six orderings. Each order gives you a way to resolve the absolute value signs. You can check them all.

Answer 2

Another way, as we are concerned only with distances between points on the real line WLOG we can shift the origin, so set $y=0$, to simplify the inequality. Now easy to check cases or use your squaring argument.

Answer 3

If you know that $|a+b|\le|a|+|b|$ holds for any real numbers $a$, $b$, $c$ (this is triangle inequality), then you can simply plug in $b=x-y$ and $c=y-z$.

You get $a+b=x-z$ and $$|x-z|\le|x-y|+y-z|.$$