TITLE: $p$-adic metric triangle inequality
QUESTION [6 upvotes]: If $d$ denotes the $p$-adic metric and $d(m,o) = {1 \over p^{t_{mo}}}$, $d(m,n) = {1 \over p^{t_{mn}}}$ and $d(n,o) = {1 \over p^{t_{no}}}$ I am trying to show the triangle inequality. But I'm not sure the proof is correct. Can you help me and correct it if it's wrong? Thank you. Here is my proof:
Have: $m-o = (m-n) + (n-o)$ where $p^{t_{mn}} | (m-n)$ and $p^{t_{no}} | (n-o)$. Then $m-o = p^{t_{mn}}k + p^{t_{no}}j$. Without loss of generality assume $t_{mn} \ge t_{no}$ so that $$m-o = p^{t_{no}}(p^{t_{mn}-t_{no}}k + j)$$ 
Particularly, the following step I am unsure of: It follows that $t_{mo} = t_{no}$ and therefore $d(m,o) = {1 \over p^{t_{mo}}} = {1 \over p^{t_{no}}} = d(n,o)$. And therefore $d(m,o) \le d(n,o) + d(m,n)$. qed

REPLY [4 votes]: You were right to worry about that last step: it isn’t quite right.
If $t_{mn}>t_{no}$, then $p^{t_{mn}-t_{no}}k+j\equiv j\not\equiv0\pmod p$, so $t_{mo}=t_{no}$. If $t_{mn}=t_{no}$, however, then $m-o=p^{t_{no}}(k+j)$, where $k+j$ may be divisible by $p$. In that case $t_{mo}>t_{no}$. However, in all cases you can safely conclude that $t_{mo}\ge t_{no}$ and hence that $d(m,o)\le d(n,o)$. 
This shows that not only is $d(m,o)\le d(n,o)+d(m,n)$, but in fact $$d(m,o)\le\max\{d(n,o),d(m,n)\}\;:$$ $d$ is a non-Archimedean metric.