TITLE: Proving this distance to be a metric?
QUESTION [6 upvotes]: $\forall x,y \in \mathbb R$ if distance $d(x,y) = \sqrt { |(x - y)|}$,
How can I prove this distance to be metric? I am stuck at triangular equality.
i.e. $ \sqrt{ |x - y| } \leq \sqrt{ | x - z|} + \sqrt{|z - y|} $
I arrived at the relation 
$\sqrt x + \sqrt y \geq \sqrt{x+y} $ 
But how the another variable $z$ comes into the picture remains question to me. Any help appreciated.

REPLY [2 votes]: By squaring we have $$\sqrt {|x-y|}\leq \sqrt {|x-z|}+\sqrt {|z-x|}\iff$$ $$ |x-y|\leq |x-z|+|z-y|+2\sqrt {|x-z|} \sqrt {|z-y|}$$ which is immediate from   $|x-y|=|(x-z)+(z-y)|\leq |x-z|+|z-x|.$