<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] You can shorten your approach and deal with the negative case at the same time, by writing
$$
(|a|+|b|)^2=a^2+2|a||b| + b^2\stackrel{(1)}=a^2+2ab+b^2=(a+b)^2\tag{*}
$$
where (1) uses the fact that $|a||b|=ab$ if $a$ and $b$ have the same sign. Now take square roots of both ends of (*).
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