How to visualize Reverse Triangle Inequality $||\mathbf{a}| − |\mathbf{b}|| ≤ |\mathbf{a} − \mathbf{b}|$?

Solution 1:

Let do this in terms of vector ...

sorry for bad drawing. I hope you will understand :D

ADDED: Here is clarification as you requested.
The length $|a| = AD = AE $ because they are radius of the same circle. And $|b| = AC$ so $||a|-|b|| = EC = AC - AE$

Solution 2:

Draw two circles as follows:

Let $z_1$ be any complex number on the larger red circle and $z_2$ be any complex number on the smaller black circle. Note that $|z_1|$ and $|z_2|$ are the radii of the red circle and black circle respectively.

$\big||z_1| - |z_2|\big|$ (purple on the real axis) is always some positive number representing the absolute value of the difference between these radii.

$|z_1 - z_2|$ (blue) is a positive number representing the distance between $z_1$ and $z_2$ (some different choices for $z_1$ and $z_2$ are shown)

As you can see no matter how you choose the points on the red and black circle, the inequality $$||z_1| - |z_2||≤|z_1 - z_2|$$ holds

The smallest distance between two points on two concentric circles is the difference between the circles's radii.

Solution 3:

The intuition is that the inequality is equality whenever there's no cancellation happening. I'll keep this in the reals but the idea generalises. Suppose that $a,b > 0$. Then \begin{equation} | |a| - |b| | = |a - b| \end{equation} whereas, if $a > 0 > b$, then \begin{align} | |a| - |b| | &= | a + b | \leq \max\{|a|, |b|\} \leq \max\{a, -b\} \leq \max\{a, -b\} + \min\{-b, a\} = a - b \\ &= |a - b| \end{align}

If you want to think more geometrically then $|a - b|$ is the size of the vector from $b$ to $a$ which is minimised when they're both pointing in the same direction and, at that minimum, is the LHS above. When they're not pointing in the same direction then $|a - b|$ is bigger but $||a| - |b||$ stays the same size (as it doesn't care about the angle between $a$ and $b$).