Proof of the Reverse Triangle Inequality

Question

Here there is my proof (quite short and easy) of a rather straightforward result. The text of this question comes from a previous question of mine, where I ended up working on a wrong inequality. Here things are fixed.

Still, I would like to know:

1. if it is sound, because absolute value always creates me some problem, and
2. if there is a shorter (neater) way to get the result (maybe without using contradiction).

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Proposition: $|d(x,y)-d(y,z)| \leq d(x,z)$.

Proof:
We proceed by cases.

• Case 1: $d(x,y)-d(y,z) \geq 0$.
  Assume by contradiction that $d(x,y)-d(y,z) > d(x,z)$. Hence, $d(x,y)> d(y,z) + d(x,z)$, contradicting the triangle inequality.

• Case 2: $d(x,y)-d(y,z) < 0$.
  Assume by contradiction that $ - d(x,y) + d(y,z) > d(x,z)$. Hence, $d(x,y) < d(y,z) - d(x,z)$. By applying the triangle inequality, we have that $d(y,x) \geq d(y,z) + d(x,z)$, which is by symmetry equivalent to $d(x,y) \geq d(y,z) + d(x,z)$, obtaining the desired contradiction. $\square$

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As always, any feedback is welcome.
Thank you for your time.

Answer 1

Your work seems right (although you could get the easier contradiction in the second case by adding $d(x,y)$), but there is an easier proof: Let $x,y,z$ be points in a metric space. Then

$$d(x,y)\leq d(x,z)+d(z,y).$$

Then,

$$d(x,y)-d(y,z)\leq d(x,z).$$

On the other hand,

$$d(y,z)\leq d(x,z)+d(x,y).$$

Then,

$$d(y,z)-d(x,y)\leq d(x,z).$$

Therefore,

$$|d(y,z)-d(x,y)|\leq d(x,z).$$

Answer 2

You don't really need to prove it by contradiction:

By the triangle inequality, $d(x,y)\le d(x,z)+d(z,y)=d(x,z)+d(y,z)$, whence $$d(x,y)-d(y,z)\le d(x,z.)$$ Similarly, $d(y,z)\le d(y,x)+d(x,z)$, whence $$d(y,z)-d(x,y)\le d(x,z),$$ and finally $$\bigl\lvert d(x,y)-d(y,z)\bigr\rvert=\max\bigl(d(x,y)-d(y,z),d(y,z)-d(x,y)\bigr)\le d(x,z).$$