Proof by cases(4)

If a and b  are real numbers.

Case 1: Assume that a >= 0, b >= 0. Then |a| = a, |b| = b, and |a - b| = a - b. So in this case
||a| - |b|| = |a - b| =< |a - b|.

Case 2: Assume that a >= 0, and b < 0. Then |a| = a and |b| = -b.

-Sub-case 1: Assume that |a| >= |b|, then 0=< a + b =< a =< a - b.
So |a + b| < |a - b| giving us

||a| - |b|| = |a + b| < |a - b|.

-Sub-case 2: Assume that |b| > |a|, then we have both 0 < -b =< a - b, and b < a + b < 0. So it follows that |a + b| < |a - b| giving us

||a| - |b|| = |a + b| < |a - b|

Case 3: Assume that a < 0, and b >= 0. Then |a| = - a and |b| = b.
-Sub-case 1: Assume that |a| >= |b|, then 0 < - a =< b - a = |a - b|, and a=< a + b =< 0. So that |a + b| =< |a - b|. It follows that

||a| - |b|| = |- a - b| = |a + b| =< |a - b|.

-Sub-case 2: Assume that |b| > |a|, then a - b =< - a - b< 0. This gives us |- a - b| =< |a - b|. So again
||a| - |b|| = |- a - b| =< |a - b|.

Case 4: Assume that a < 0, and b < 0. Then |a| = - a and |b| = - b.
Then in this case
||a| - |b|| = |- a + b| = |a - b| =< |a - b|.

So in all cases ||a| - |b|| =< |a - b|

QED