TITLE: Proving that a metric is non-negative
QUESTION [6 upvotes]: I wanted to try a problem, where I need to prove that the non-negativity of a metric follows from the following axioms: 

For a metric $d$ in some space $X$, we have for $x,y,z\in X$
$$(1)\;\;\;\;d(x,y)=0\;\;\text{iff}\;\;\;x=y$$
  $$(2)\;\;\;\;\;\;\;\;\;d(x,y)=d(y,x)$$ $$(3)\;\;\;\;d(x,y)\leq
 d(x,z)+d(z,y)$$

My question is that, is this a sufficient proof? :
From triangle inequality we have: 
$$d(x,y)-d(z,y)\leq d(x,z)$$
By setting $y=x$ it follows from $(1)$ and $(2)$ that: 
$$-d(x,z)\leq d(x,z)$$
and this can only be true if $$d(x,z)\geq0$$

REPLY [7 votes]: $$0 = d(x,x) \leq d(x,y) + d(y,x) = 2 d(x,y) $$
and so
$$ d(x,y) \geq 0 $$