Proving the triangle inequality property of a metric I have to show that $d(x,y) = \sqrt{|y-x|}$ on $\mathbb{R}$ is a metric. The first two properties were fairly trivial, however I'm having issues showing that $d(x,z) \leq d(x,y) + d(y,z)$. I tried this approach: $d(x,z) = \sqrt{|z-x|}  = \sqrt{|z-y+y-x|} \leq \sqrt{|z-y| + |y-x|}$, but it doesn't seem to be getting me anywhere.

Hint: For $a,b\ge0$
$$\sqrt{a+b}\le\sqrt{a}+\sqrt{b}$$
as squaring  each side we get
$$
a+b\le a+b+2\sqrt{ab}
$$
