Triangle inequality for subtraction?

It's sometimes called the reverse triangle inequality. The proper form is $$\left| a - b \right| \ge \big||a| - |b|\big|$$ For the proof, consider $$|a| = |a - b + b| \le |a - b| + |b|$$ $$|b| = |b - a + a| \le |a - b| + |a|$$ so that we have $$-|a-b|\le|a|-|b| \le |a - b|$$

No. For example, $|(-2)-3|=5>|-2|-|3|=-1.$

I think you're thinking of $||a|-|b||\le |a- b|.$

The length of any side of a triangle is greater than the absolute difference of the lengths of the other two sides:

$$||a|-|b||\leq |a-b|$$

Here is a proof:

$$|a+(b-a)|\leq |a|+|b-a|$$

and,

(1) $$|a-b|\geq |a|-|b|$$

Interchanging $a$ and $b$, we get also

(2) $$|a-b|\geq |b|-|a|$$

Combining (1) and (2) we get our desired result.