Absolute value property (proof)

I want to prove $$\lvert a \rvert \geq b \iff a\leq-b \quad \text{ or } \quad a\geq b,$$ for $b>0$.

Case 1: $a>0$, $\lvert a \rvert=a$ so $$a\geq b$$

Case 2: $a=0$, $\lvert a \rvert=0$ so $$0\geq b, \quad \textbf{not possible because }\mathbf{b>0?}$$ Case 3. $a<0$, $\lvert a \rvert =-a$ so $$-a\geq b$$

Summary:

$a>0: a\geq b \iff a-b\geq 0$ $a<0: -a\geq b \iff b+a\leq 0$

This gives \begin{align} b+a&\leq 0 \leq a-b &\iff \\ b+a &\leq a-b &\iff \\ 2b &\leq 0 &\iff \\ b &\leq 0 \quad \textbf{What?} \end{align}

Questions:

Why is case 2 a contradiction?

Why can I not combine the inequalities (final step?

• You can't combine the inequalities because $a$ can't be both positive and negative, that's all. – Bernard Apr 9 '17 at 21:26

I'll answer each of your questions in turn.$\newcommand{\OR}{\ \ {\rm{\small{OR}}}\ \ }$
In case 2, you've discovered that the left-hand side of the equivalency (that is, the statement $|a| \geq b$) implies a contradiction if $a$ is assumed to be zero and $b$ is supposed to be positive. This means that the left-hand statement is false (when $a=0$), but this doesn't mean that your entire statement-to-be-proved is false, because you are trying to prove an equivalency, not just one particular statement that's part of the equivalency. Remember, to say "$p \iff q$" is to say that the statements $p$ and $q$ have the same truth value: are either both true or both false. "$|a| \geq b$" may be false for $a = 0$, but so is the statement "$a \leq -b \OR a \geq b$" because with $a=0$ both of these also imply $b \leq 0$. Lies are equivalent to lies, and hence your entire equivalency is true for $a = 0$.
As for your second question, you can't combine the inequalities in the way you did. As Bernard pointed out, the two inequalities you combine come from different cases: $a-b \geq 0$ if $a>0$, and $b+a \leq 0$ if $a<0$. Therefore you can't assume that the two inequalities are both true at the same time which is necessary before you can combine them. It turns out that whenever one inequality is true, the other one must be false. Hence you can never combine them in any situation.
But even if you could combine those inequalities, I don't think it would lead anywhere useful. Once you've got your three previous cases finished, I would then attempt to prove the converse by assuming first that $a \leq -b$ and proving that must mean $|a| \geq b$, and then independently assuming $a \geq b$ and then proving $|a| \geq b$. Once you do that (and complete the proofs for your original case 2 and case 3), you will have completed the proof of $$|a| \geq b \iff a \leq -b \OR a \geq b$$