I didn’t finish a careful proof of the triangle inequality in class, so I am presenting a more polished argument here. I abandoned the text’s argument in favor of what I hope is a more intuitive (albeit longer) proof.

I’ve also had a number of requests for hints on a couple homework problems, specifically numbers 3.7 and 3.8. So here are my comments on those problems:

Problem 3.7 This problem asks us to prove that \(|b| < a\) if and only if \(- a < b < a\), and some variations on this result. Most often, it is easiest to prove "if and only if" statements in two parts: first prove that if \(|b| < a\) then \(-a < b < a\); then prove that if \(-a < b < a\) then \(|b| < a\). Notice that the statement \( a < b \) is equivalent to \(a \leq b\) and \(a \neq b\). Therefore, you can use all of the order axioms for the strict inequality (\(<\)) except that O2 doesn't apply.
You can prove the first implication by breaking the problem up into the cases where \(b\) is positive and negative. Also, you can assume that \(a > 0\), for otherwise the assumption \(|b| < a\) is impossible. If \(b \geq 0\) then \(|b| = b\) so that \(|b| < a \Rightarrow b < a\). On the other hand, we get \(-a < 0 < b\). Combining this with the previous inequality gives \(-a < b < a\), as desired. The case where \(b\) is negative is similar.
For the opposite implication (assuming \(-a < b < a\)), it is again helpful to break the proof up into cases depending on the sign of \(b\). For example, if \(b \geq 0\), then \(|b| = b < a\), which does it. If \(b \leq 0\), you'll have to appeal to the assumed inequality \(-a < b\).
Part (b) asks you to prove that \(|a - b| < c\) if and only if \(b - c < a < b + c\), and (c) asks you to prove the same thing replacing each \(<\) with \(\leq\). You can prove (b) by appealing to (a): just use \(a - b\) for \(b\) and \(c\) for \(a\) in part (a) and see what you get. For part (c), you can just argue that the same inequality as (a) holds with \(\leq\), so the proof of (c) is essentially the same as the proof of (b). Don't worry about going through all the details again -- just convince me that you understand what is going on in you write-ups.
Problem 3.8 In this problem, you are asked to prove that for \(a, b \in \mathbf{R}\) if \(a \leq b_1\) for every \(b_1 > b\), then \(a \leq b\).

I think the easiest way to approach this problem is to do a proof by contradiction. The idea is that in order to prove that a statement is true, you prove derive a contradiction from its negation. The negation of the statement we are trying to prove is, “for all \(b_1 > b\), \(a \leq b_1\) and \(a > b\).”

If you think about what this statement is saying, it should be clear that it cannot possibly be true. For example, if \(a > b\), then we can choose \(b_1 = (a + b) / 2\) so that \(b < b_1 < a\). But this choice of \(b_1\) contradicts the assumption that for all \(b_1 > b\) \(a \leq b_1\). Since the negation of the statement we are trying to prove leads to a contradiction, the statement must be true.