Show that $d(x, y) \ge 0$ whatever $x, y \in X$

Question

Let $X$ be a set and $d : X \times X \to \mathbb R$ be a function satisfying the following conditions:

$d(x, x) = 0$ whatever $x \in X$;

$d(x, y) = d(y, x)$ whatever $x, y \in X$;

$d(x, z) \le d(x, y) + d(y, z)$ whatever $x, y, z \in X$.

Show that $d(x, y) \ge 0$ whatever $x, y \in X$.

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My attempt

$d(x, y) \le d(x, y) + d(x, y)$

$d(x, y) \le d(x, y) + d(x, y)$

$d(x, y) - d(x,y) \le d(x, y)$

$0 \le d(x, y)$

I know it shouldn't be too difficult but I'm not sure about my answer, is that correct?

Answer 1

$0=d(x,x)\le d(x,y)+d(y,x)=d(x,y)+d(x,y)=2d(x,y)$

where all properties 1,2 and 3 have been used.

Answer 2

You don't say how you establish $$ d(x,y)≤d(x,y)+d(x,y) $$ but once you do have this, the following steps are valid. In general, not all quantities $r \in \mathbb{R}$ satisfy $r \le r + r$; for example $r = -1$. So we need to do a bit more work to get an inequality like that, from the information given.

From the axioms, the only one with a "$\le$" in it is the third. We should look to see if there is a choice of $x$, $y$ and $z$ that will let us get something of the form "$0 \le \dots$", which is what we need. There is a $0$ in the first axiom, which we can reach by letting $x = z$ in axiom (3), like this: $$ d(x, x) \le d(x, y) + d(y, x) $$ Now, $d(x, x) = 0$, and $d(x, y) = d(y, x)$, so you should be able to take it from here.