Prove: Use the triangle inequality to prove that for all $x, y, | |x| − |y| | ≤ |x − y|$
Proof:
If $x ≥ 0$ and $y ≥ 0$, then both sides of the inequality are the same.
Also if $x ≤ 0$ and $y ≤ 0$ then both sides of the inequality are the same.
Assume $x ≥ 0$ and $y ≤ 0$ then $|x| = x$, and $|y| = −y$, and so:
$||x|| = ||x − y + y|| ≤ ||x − y + y|| ⇒ ||x−y||≤||x − y||$
and
$||y|| = ||x − y − x|| ≤ ||x − y|| + ||x|| ⇒ −(||x||−||y||) ≤ ||x − y||$
How do I approach this? Is this a good start?
Update:
I feel the following approach is a little bit more clear.
Proof:
Case $x≥0$ and $y≥0$: We have both sides equal each-other
Case $x≤0$ and $y≤0$: We also have both sides equal each-other
Case $x≥0$ and $y≤0$: Then $|x|=x$, and $|y|=−y$, and so $||x|| = ||x − y + y|| ≤ ||x − y + y|| ⇒ ||x−y||≤||x − y||$
Case $x≤0$ and $y≥0$: Then $|x|=-x$, and $|y|=y$, and so $||y|| = ||x − y − x|| ≤ ||x − y|| + ||x|| ⇒ −(||x||−||y||) ≤ ||x − y||$
QED
