# Showing $\left| (|a|-|b|)-(|c|-|d|) \right| \le \left| (a-b)-(c-d) \right|$

I am trying to show that for any real $$a,b,c,d$$ we have

$$\left| (|a|-|b|)-(|c|-|d|) \right| \le \left| (a-b)-(c-d) \right|$$

At a glance, it seems like this cannot be true, but I haven't been able to come up with a counter example. Furthermore, it is not to hard to establish that

$$\left| |a-b|-|c-d| \right| \le \left| (a-b)-(c-d) \right|.$$

For example by the (reverse) triangle inequality we have either $$\left| |a-b|-|c-d| \right| = |a-b|-|c-d| \le |(a-b)-(c-d)|$$ or $$\left| |a-b|-|c-d| \right| = |c-d|-|a-b| \le |(c-d)-(a-b)| = |(a-b)-(c-d)| .$$

And it seems much more believable that possibly $$\left| (|a|-|b|)-(|c|-|d|) \right| \le \left| |a-b|-|c-d| \right|$$ could hold, establishing the desired result.

A counterexample is $$a=1,b=c=0,d=-1$$.
• Dang. I knew it would be obvious and was reluctant to post for that reason. Thank you nonetheless. I wonder then what additional restrictions could make the above true. All nonzero? All $a,b,c,d$ distinct? Not really at all? In any case I think under too many more restrictions the inequality above loses all usefulness anyway. Jan 26, 2022 at 17:13