Double absolute value proof: $||a|-|b||\le |a-b|$ I guess, I know how to solve inequalities with absolute value, but I have problems with this one.
$||a|-|b||\le |a-b|$
$a,b\in \mathbb{R}$
I tried to solve the inequation like this:
case1: $a>0$ 
case2: $a<0$ 
I started with case 1. than we have two possibilities $b<0$ and $b>0$ 
Firstly I took $b<0$ and I had another two possibilities
$|a+b|>0$
$|a+b|<0$
I take $|a+b|>0$
and in this case I had two possibilites according to to right side of the inequation
$|a-b|>0 ........... a+b\le a-b ......b\le-b$
$|a-b|<0..........a+b\le -a+b .......a\le -a$
I can do the same with other possibilites, but how do I know, if my solution (or even method) is right? there are so many conditions, that I am lost in them.
Thank you for your time.
 A: Instead of considering signs of $a$ and $b$ separately, you can consider their relative signs: if they have the same sign (or one is zero), then the two sides are equal, while if they have different (nonzero) signs, the right hand side is bigger (equal to sum of the absolute values).
If this is not obvious to you, then you can simplify a bit by assuming that $a$ is nonnegative: you can do that because flipping signs of $a$ and $b$ both does not change either side.
A: $$
||a|-|b||\le |a-b|
$$
iff
$$
||a|-|b||^2\le |a-b|^2
$$
iff
$$
a^2 -2|ab| +b^2 \leq a^2-2ab+b^2
$$
iff
$$
ab \leq |ab|,
$$
and the last inequality is always true.
A: There is a simpler solution.
First, notice that the inequality you want to show is equivalent to
$$|a| - |b|\leq |a-b| \text{ and } |b| - |a|\leq |a-b|$$
Let's see how to show the first one.
We have $|a| = |(a-b) + b|$. Using the triangular ineaquality, this implies that $|a|\leq |a-b| + |b|$, thus $|a|-|b|\leq |a-b|$. You can show the other inequality exactly the same way.
