Triangle inequality proof by cases

In Introduction to Analysis by Trench, there is a proof by cases for the triangle inequality, stated as follows:

$(a),\ if\ a\geq 0, and\ b\geq 0,\ then\ a+b \geq0\ , So\ |a+b| = a+b= |a|+|b|$ $(b),\ if\ a\leq 0, and\ b\leq 0,\ then\ a+b \leq0\ , So\ |a+b| = -a+(-b)= |a|+|b|$ $(c),\ if\ a\leq 0, and\ b\geq 0,\ then\ a+b = -|a|+|b| , So\ |a+b| = |-|a|+|b||\leq |a|+|b|$ $(d),\ if\ a\geq 0, and\ b\leq 0,\ then\ a+b = |a|-|b| , So\ |a+b| = ||a|-|b||\leq |a|+|b|$

What I wanted to understand is the last two statements (c) and (d). I tried to sketch it as follows so it makes sense if that's true: $||a|-|b||=||b|-|a||\leq ||b|+|a||=|b|+|a|$

Is it the way it is done to show the (c) and (d) are true?

You can also obtain the triangle inequality by using $\forall a\;(a\leq |a|)$ and $\forall a \;(|$-$a|=|a|)$ and $\forall a\;(|a|\in \{a,\;$-$a\},$ as follows:

(i). $a+b\leq |a|+b\leq |a|+|b|.$

(ii). -$(a+b)=($-$a)+($-$b)\leq |$-$a|+($-$b)\leq |$-$a|+|$-$b|=|a|+|b|.$

(iii). Now $|a+b|\in \{a+b,\;$-$(a+b)\}$ while $a+b$ and -$(a+b)$ are each $\leq |a|+|b|.$

Yes, that is ok, although the author is getting directly the desired result using that $a = -|a|$ for $a \leq 0$ and $-|a| \leq |a|$ for all $a$.

• I don't think he is using such. Because he moves to So, where ||b|−|a|| ≤ |a|+|b| and not |b|−|a| ≤ |a|+|b| so that he can use such properties. – Papa Sep 12 '18 at 16:52
• Sorry I don't understand what you mean. In my opinion, he/she is making explicit use of those properties – Javi Sep 13 '18 at 1:47
• How can you infer ||b|−|a|| ≤ |a|+|b| from only ||b|−|a|| using such properties? – Papa Sep 13 '18 at 5:25
• By the 2nd property $-|a| \leq |a|$ then $|b|+(-|a|)=|b|-|a| \leq |b|+|a|$ so if $|b|-|a|\geq0$ then $||b|-|a|| = |b|-|a| \leq |b|+|a|$. On the other hand, if $|b|-|a|<0$ then $|a|-|b|>0$ and proceeding the same as before $||b|-|a|| = ||a|-|b|| \leq |a|+|b|$ – Javi Sep 13 '18 at 13:34
• Thanks Javi, that was clear! – Papa Sep 13 '18 at 13:49