Spivak absolute value Problem: abs(abs(a + b) + abs(c) - abs(a + b + c)) [SOLVED]

Question

Currently I'm working on Spivak's Calculus but I'm stuck at the following example, where one should drop at least one pair of absolute value signs:

$\mid(\mid a+b\mid + \mid c \mid -\mid a+b+c \mid) \mid$

The solution he's giving looks like this:

$\mid a+b\mid + \mid c \mid - \mid a+b+c \mid$

As I was working on this problem I thought I came up with something true in the form of:

$\mid a \mid = \mid $-$(\mid a \mid)\mid$

Therfore I thought I can just (inversly) distribute $ -1$ to get:

$\mid-(-\mid a+b\mid - \mid c \mid +\mid a+b+c \mid) \mid$ = $-\mid a+b\mid - \mid c \mid +\mid a+b+c \mid$

and then drop the outmost $abs$ $sign$. But apparently that's not working. Can somebody please help me understand where I went wrong with this (equation 3) and explain to me how the given solution can be obtained?

Answer 1

Notice that the triangle inequality gives you that

$$\lvert a+b+c\rvert\leq \lvert a+b\rvert +\lvert c\rvert,$$

which means that

$$\lvert a+b\rvert +\lvert c\rvert-\lvert a+b+c\rvert\geq 0.$$

We can thus remove the absolute value to get that

$$\bigl\lvert\lvert a+b\rvert +\lvert c\rvert-\lvert a+b+c\rvert\bigr\rvert=\lvert a+b\rvert +\lvert c\rvert-\lvert a+b+c\rvert.$$

Answer 2

You have $$|a+b+c|\leq |a+b|+|c|$$ by triangular inequality so $$|a+b|+|c|-|a+b+c|\geq 0.$$ Therefore, $$||a+b|+|c|-|a+b+c||=|a+b|+|c|-|a+b+c|.$$