Proof of triangle inequality for $d(x,y)=\sqrt{\lvert x-y\rvert}$ There is this problem that says: show that $d(x,y)=\sqrt{\lvert x-y\rvert}$ is a distance function on $\mathbb{R}$, and I am unable to proof the triangle inequality for this? any suggestion I look forward to. 
thanks.

Hint:
We already have the usual triangle inequality $\lvert x-y\rvert\leq\lvert x-z\rvert+\lvert z-y\rvert$. This tells you that
$$
\sqrt{\lvert x-y\rvert}\leq\sqrt{\lvert x-z\rvert+\lvert z-y\rvert}.
$$
From here, it sufficies to prove that $\sqrt{a+b}\leq\sqrt{a}+\sqrt{b}$ for non-negative $a,b$.
