I am reading "Topological Spaces: Examples and Exercises" by Tetsuro Kawasaki (in Japanese).
There is the following exercise in this book:
Definition:
Let $X$ be a set.
Let $d$ be a function $X\times X\to\mathbb{R}$ which satisfies (D1), (D2), (D3), (D4).
(D1) $d(x,y)\ge 0$.
(D2) $d(x,y)=0\Leftrightarrow x=y$.
(D3) $d(x,y)=d(y,x)$.
(D4) $d(x,y)+d(y,z)\ge d(x,z)$.
Exercise 3.13:
Show that (D1) follows from (D2), (D3), (D4).
The author's answer:
$d(x,y)+d(y,x)\ge d(x,x)$ by (D4).
$d(x,x)=0$ by (D2).
So, $d(x,y)+d(y,x)\ge 0$.
By (D3), $2d(x,y)\ge 0$.
$\therefore d(x,y)\ge 0$.
The author used (D2),(D3),(D4) to show (D1), but I think we don't need (D2) to show (D1).
My answer:
Assume that there exists $(x,y)\in X\times X$ such that $d(x,y) < 0$.
Then, by (D4),
$d(x,y)\ge d(x,z)-d(y,z)$ and
$d(y,x)\ge d(y,z)-d(x,z)$.
By (D3),
$d(x,y)\ge d(y,z)-d(x,z)$.
Since $d(x,y)<0$, both $d(y,z)>d(x,z)$ and $d(x,z)>d(y,z)$ hold.
This is a contradiction.
Am I correct or not?
Yes you're right, I think you can also show it this way: $$ d(x,y) \geq d(x,z) - d(y,z) \text{ } \text{ } \text{[From your answer]}$$ $$ d(x,y) \leq d(x,z) + d(z,y) \text{ } \text{ } \text{[Standard Triangle]}$$ and so subtracting $d(x,z)$ gives $$ -d(y,z) \leq d(z,y)$$ but since symmetry holds this can only work if $d(\cdot,\cdot)\geq0$.
(This is pretty much similar to your proof)