Using the triangle inequality to show that disjoint balls can be formed around distinct points I want to start by saying that I'm familiar with the canonical proof of this result. I would like to prove it using a specific lemma:
Let $x, y, z \in \mathbb{R}$. Let $k \in \mathbb{R}$ such that $d(x, z) < k < d(x, y)$. Then $d(x, y) - k < d(y, z)$.
I can provide a quick example to show my intuition. If $x$ and $y$ are $10$ units apart, and $z$ is less than $7$ units away from $x$, then $z$ must be more than $3$ units away from $y$. (I wanted to prove a more general case than just using $\frac{d(x,y)}{2}$.)
I can only assume that this lemma will also need the triangle inequality. But I'm not sure how to actually prove it. Writing proofs with the triangle inequality is still a bit uncomfortable for me and I'm at my wit's end. Could someone help me prove my lemma? It seems that a proof by contradiction is the way to go, but that's about as far as I've gotten, haha.
 A: One of the confusing things about the triangle inequality with three points $x,y,z$ is that there are actually three triangle inequalities:
$$(1) \quad d(x,z) \le d(x,y) + d(y,z)
$$
$$(2) \quad d(y,x) \le d(y,z) + d(z,x)
$$
$$(3) \quad d(z,y) \le d(z,x) + d(x,y)
$$
All three of these are true. 
But, when you are working with three particular named points $x,y,z$ and you want to derive something specific, maybe only one of the three is useful. You may need to choose the correct one from (1), (2), or (3) in order to obtain useful information.
Which one do you pick? 
Look at what you are trying to prove, namely something of the form $d(z,y) > \square$. Which one of (1), (2) or (3) gives you an inequality of that form? Answer: (1) and (2) but not (3). Using (1) together with the symmetry law $d(z,y) = d(y,z)$, one has
$$(1') \quad d(z,y) \ge d(x,z) - d(x,y)
$$
Using (2) one has
$$(2') \quad d(y,z) \ge d(x,y) - d(x,z)
$$
Both of these are true.
But notice: $(2')$ has form closest to what you are actually trying to prove, namely that
$$d(y,z) > d(x,y) - k
$$
Finally, use what you are given and combine it with $(2')$: if $k > d(x,z)$ then $-k < -d(x,z)$ and so $-d(x,z) > -k$ and therefore
$$d(y,z) \ge d(x,y) - d(x,z) > d(x,y) - k
$$
Your final cleanly written up proof can now simply ignore (1) and (3) and use only (2).
A: By triangle inequality,
$d(x,y) \le d(x,z) + d(y,z).$
But $d(x,z) \lt k.$
Hence $d(x,y) \lt k + d(y,z) \Rightarrow d(x,y)-k \lt d(y,z).$
