Proving $|a+b|\le |a|+|b|$ from $-|a|\le a \le |a|$

Question

In Spivak Calculus chapter 1, question no. 14, it is asked to prove the aforementioned inequality. However, the way I proved it is unnecessarily long. Can someone critique it for me, and mention an alternate shorter proof method if any?
My proof:
$-|a|\le a \le |a| \implies -|a+b|\le a+b \le |a+b|$

Case 1: $a+b\ge 0$
$-|a+b|\le a+b \le |a+b| \implies -(a+b)\le a+b \le (a+b) \implies -a-b\le a+b \le a+b $

Subcase 1: $a\ge 0 ,b\ge 0$:

$-a-b\le a+b \le a+b \implies -|a|-|b|\le a+b \le |a|+|b| \implies -(|a|+|b|)\le a+b \le (|a|+|b|) \\ \implies |a+b| \le |a|+|b|$

Subcase 2: $a\ge 0 ,b< 0$:

$-a-b\le a+b \le a+b \implies -|a|+|b|\le a+b \le |a|-|b| \implies |a+b| \le |a|-|b| \implies |a+b| \le |a|+|b|$

Subcase 3: $a<0, b \ge 0$ Since the inequality is symmetric in $a$ and $b$, Subcase 2 applies on this one too.

Subcase 4: $a < 0,b < 0$: Not applicable

I analyzed similarly the four subcases for case 2: $a+b \le 0$ but you can see the proof is getting too long.

Answer 1

Adding

$$-|a|\le a\le |a| $$ and

$$-|b|\le b\le |b|.$$

you get

$$-(|a|+|b|)\le a+b\le|a|+ |b|,$$

which is

$$-(a+b)\le|a|+|b|\land a+b\le|a|+|b|$$ or $$|a+b|\le|a|+|b|.$$

Answer 2

Consider cases of $a,b$ being same signs or opposite signs. and note $x = \pm |x|$ (and $|x| =\pm x$). And then if they are opposite signs have as subcases $|b| \le |a|$ or $|b| > |a|$.

Case 1: $a,b$ are either both $< 0$ or both $\ge 0$.

Then $a+b = \pm |a| \pm |b| = \pm (|a| + |b|)$ so $|a+b| = |\pm(|a|+|b|)| = |a| + |b|$.

Case 2: $a,b$ are opposite signs, i.e. either $a < 0 \le b$ or $b < 0 \le a$.

Then $a+b = \pm |a| \mp |b| = \pm(|a|-|b|)=\mp(|b| -|a|)$

Subcase 2a: $|a| \ge |b|$ so $|a|-|b| \ge 0$

Then $|a+b| = |\pm(|a|-|b|)|= ||a|-|b|| = |a|-|b| \le |a| \le |a| + |b|$.

Subcase 2b: $|a| < |b|$ so $|a|-|b| < 0$ and $|b|-|a| > 0$. so then $|a+b| = |\pm (|a|-|b|)|=|\mp(|b|-|a|)|=|b|-|a| < |b| \le |a| + |b|$.