<s>[INST] Show that if a and b have the same sign then |a| + |b| = |a + b| We start with  a, b > 0.
We know that $|a| = \sqrt{a^2}$, $|b| = \sqrt{b^2}$ and $|a+b| = \sqrt{(a+b)^2}$
We do the following :
$|a| + |b| = \sqrt{a^2} + \sqrt{b^2}$
$=a + b$
So : $|a| + |b| = a+b$
We take this result : 
$|a| + |b| = a+b$
$(|a| + |b|)^2 = (a+b)^2$
$\sqrt{(|a| + |b|)^2} = \sqrt{(a+b)^2}$
$a+b =|a+b|$
Thus : |a+b|= |a| + |b|
For the case where a, b < 0, we use the same technique but with minuses before the constants Opinion ? 
 [/INST] If you want to use the characteritic that $|a| = \sqrt{a^2}$ you certainly can.
$(|a|+|b|)^2 = a^2 + b^2 + 2 |ab|$
if |ab| = ab then
$(|a|+|b|)^2 = (a+b)^2$
$||a|+|b|| = |a+b|$
$|a|+|b| = |a+b|$
</s>