Spivak Calculus 3rd ed. $|a + b| \leq |a| + |b|$ I'm working through the first chapter of Michael Spivak's Calculus 3rd ed.  
Towards the end of the chapter he proves $ |a + b| ≤ |a| + |b| $ using the observation that $|a|= \sqrt{ a^2 }$ when  $a$ is $ ≥ 0 $ .
$ |a + b| ≤ |a| + |b| $
$$ (|a + b|)^2 = (a + b)^2 $$ $$= a^2 + 2ab + b^2 $$
$$ ≤ a^2 + 2|a|  |b| + b^2 $$ 
$$ = |a|^2 + 2|a|  |b| + |b|^2 $$
$$ = (|a| + |b|)^2 $$
I am unsure about what's going on with the equality sign. How does it go from $=$ to $≤$ on line 3 when a and b are changed to their absolute value and back to $=$ again on line 4 when $a^2$ and $b^2$ are changed to their absolute values?

Let $a=r_1(\cos A+i\sin A)$  and  $b=r_2(\cos B+i\sin B)$ 
So, $|a|=r_1$ and $|b|=r_2$
Now, $|a+b|=\sqrt{(r_1\cos A + r_2\cos B)^2+(r_1\sin A + r_2\sin B)^2}$
$=\sqrt{r_1^2+r_2^2+2r_1r_2\cos(A-B)}$
$≤\sqrt{r_1^2+r_2^2+2r_1r_2}\ $    as $\cos(A-B)≤1$ for real A,B
$=r_1+r_2=|a|+|b|$
Also observe,  $|a+b|=\sqrt{r_1^2+r_2^2+2r_1r_2\cos(A-B)}≥\sqrt{r_1^2+r_2^2-2r_1r_2}=||r_2|-|r_1||$   as $\cos(A-B)≥-1$ for real A,B
$|a+b|≥||b|-|a||$
