Triangular inequality for metric in $\mathbb{R}/\mathbb{Z}$

Question

I'm having trouble in verifying the triangular inequality for the metric $d([x],\,[y]) \doteq \displaystyle\min_{n\,\in\,\mathbb{Z}} \left|x - y + n\right|$ in $\mathbb{R}/\mathbb{Z}$, because of this minimum, I can't just assume that the minimum is compatible with the usual triangular inequality in $\mathbb{R}$.

My approach was the following: $$d([x],\,[z]) = \displaystyle\min_{n\,\in\,\mathbb{Z}} \left|x - z + n\right| \\ = \displaystyle\min_{n\,\in\,\mathbb{Z}} \left|(x - y) - (z - y) + n\right|.$$ Writing $n = p - q, p,\,q \in \mathbb{Z}$, $$\displaystyle\min_{p,\,q\,\in\,\mathbb{Z}} \left|(x - y) - (z - y) + (p - q)\right| \\ = \displaystyle\min_{p,\,q\,\in\,\mathbb{Z}} \left|(x - y + p) - (z - y + q)\right|,$$ now those absolute values are what we want, for $d([x],\,[y])$ and $d([y],\,[z])$, but I don't know if the usual triangular inequality works well with this minimum over $p$ and $q$. How should I proceed?

Any help appreciated!

Answer 1

I am omitting the $\mathbb{N}$s below:

$\min_n |x-y+n| = \min_{n,m} |x-y+n-m| \le |x-y+a-b| \le |x-a|+|y-b|$ for any $a,b$.

Now take the $\min$ over $a$ and then the $\min$ over $b$ to get the desired result.