<s>[INST] Show that if $a$ and $b$ have the same sign then $|a + b| = |a| + |b| $, and if $a$ and $b$ have opposite signs then $|a+b| < |a| + |b|$ I'm considering the different cases for $a$ and $b$
Case 1) $ a\geq 0$ and $ b\geq 0$
Given both terms are positive, $ a + b \geq 0 $
$$
|a+b| = a + b = |a| + |b|\\
$$
Case 2) $ a< 0$ and $ b< 0$
Given both terms are negative, $ a + b < 0 $
$$
|a+b| = -(a + b) = -a -b = |a| + |b|\\
$$
Case 3) $ a> 0$ and $ b< 0$
Case 3a) $ a + b \geq 0 $
Knowing that $ b < -b $ because $ b < 0 < -b $:
$$
|a+b| = a + b < a - b = |a| + |b|
$$
Case 3b) $ a + b \leq 0 $
Knowing that $ -a < a $ because $ -a < 0 < a $:
$$
|a+b| = -(a + b) = -a - b < a - b = |a| + |b|
$$
Case 4) $ a< 0$ and $ b> 0$
Case 4a) $ a + b \geq 0 $
Knowing that $ a < -a $ because $ a < 0 < -a $:
$$
|a+b| = a + b < -a + b = |a| + |b|
$$
Case 4b) $ a + b \leq 0 $
Knowing that $ -b < b $ because $ -b < 0 < b $:
$$
|a+b| = -(a + b) = -a -b < -a + b = |a| + |b|
$$
Is my proof correct?
I feel like it's somewhat incomplete as I'm not considering when either $a$ or $b$ are equal to zero in the cases 3 and 4.
 [/INST] Dealing with absolute values, the general strategy is: if you can avoid the cases method, do it.
Here , we have a typical example of this strategy: as both expressions are non-negative numbers, we compare them by comparing their squares, using that $|x|^2=x^2\,$:
$$|a+b|^2=a^2+2ab+b^2,\enspace \text{ and }\enspace (|a|+|b|)^2=a^2+2|a||b|+b^2,$$
so that comparing $|a+b|$ and $|a|+|b|$ amounts to comparing $ab$ and $|a||b|=|ab|$, namely
$$|a+b| <|a|+|b|\iff  ab<|ab| \iff ab<0,$$
i.e., in ordinary language $a$ and $b$ have different signs.
</s>