Confirming an answer using the Triangle Inequality for complex numbers

Question

I was wondering if someone can point me in the right direction in solving the following complex number problem. The problem is listed below.

The variable complex number z satisfies |z − 2 − i| = 1. Find the maximum and minimum values of |z − 3i|.

Below is my diagram. By using the diagram one can clearly see the result is

$$ 2\sqrt{2} - 1 \leq |z − 3i| \leq 2\sqrt{2} + 1 $$

Where $z_1$ is the minimum value and $z_2$ is the maximum value.

It's the next part that confuses me.

Confirm your answers to the previous part by using the triangle inequality.

$$ \big||z| − |w| \big| ≤ |z + w| ≤ |z| + |w| $$

So... what do I choose for $w$ and $z$?

Any help appreciated.

UPDATE:

Thanks to dxiv below is my solution.

We need to find:

$$ min-value \leq |z - 3i| \leq max-value $$

We also know that $ |z − 2 − i| = 1 $, so can we express $ |z - 3i| $ using $ z − 2 − i $

Thanks to dxiv $ z − 3i = (z − 2 − i )+( 2 − 2i ) $.

Thus

$$ min-value \leq | (z − 2 − i ) + ( 2 − 2i ) | \leq max-value $$

This is in the form

$$ \big||z| − |w| \big| ≤ |z + w| ≤ |z| + |w| $$

$$ \big||z − 2 − i | − |2 − 2i| \big| ≤ |z - 3i| ≤ |z − 2 − i | + | 2 − 2i| $$

$$ \big| 1 − 2\sqrt{2} \big| ≤ |z - 3i| ≤ 1 + 2\sqrt{2} $$

$$ 2\sqrt{2} - 1 \leq |z − 3i| \leq 2\sqrt{2} + 1$$

Thanks again.