Q: Does $|f(x)|≥g(x)$ imply $-g(x)≥f(x)≥g(x)$? Since for ever I thought that $|f(x)|≥g(x)\implies -g(x)≥f(x)≥g(x)$, but is it actually true? And same with $>$ instead of $≥$, does it hold? Because I have been looking for research on google and on math exchange and I was told that no. But then I came to this problem:$|x-5\pi|≥4\pi$ which if you apply so, it does seem to work? I think at least if I'm not mistaken it implies $$-4\pi≥x-5\pi≥4\pi$$ $$\pi≥x≥9\pi$$ which in interval notation it means $$x\in (-\infty, \pi]\cup [9\pi, \infty)$$ which is correct from Wolfram Alpha. What's going on here?

A: When you write
$$ \pi \geq x \geq 9\pi $$
it means
$\pi \geq x$ and $x \geq 9\pi$
(simultaneously, for the same value of $x$),
which by transitivity of $\geq$ implies that $\pi \geq 9\pi.$
That is not correct, of course, so the original set of inequalities is a mistake.
What actually is true is that
$\pi \geq x$ or $x\geq 9\pi.$
That is the answer you finally wrote and the answer Wolfram Alpha gave you.
So you have the right result but a very dubious method that relies on thinking something very different from what you are writing.

Regarding your general question,
$\lvert f(x)\rvert \geq g(x)$
implies that $f(x)\leq-g(x)$
or $g(x)\leq f(x).$
Very rarely will both inequalities be true at the same time (for the same value of $x$ — perhaps you can deduce what the one exception is)
and if you have $>$ instead of $\geq$
then the two inequalities will never both be true at the same time.

A: This is a common mistake. People are used to replacing "$|a|\leq b$" with "$-b\leq a \leq b$" and so assume you can do the same thing with the inequality reversed, i.e., that you can replace "$|a| \geq b$" with "$-b \geq a \geq b$". This is just plain wrong.
To find the correct statement, let's analyze the statement more precisely.

*

*"$|a|\geq b$" is equivalent to "not ($|a|< b$)".


*That is equivalent to "not ($-b<a<b$)".


*The compound inequality "$-b<a<b$" is shorthand for "$-b<a$ and $a<b$".


*So "not ($-b<a<b$)" is equivalent to "not ($-b<a$ and $a<b$)".


*By DeMorgan's laws, that is equivalent to "not ($-b<a$) or not ($a<b$)".


*Again by DeMorgan, that is equivalent to "$-b \geq a$ or $a \geq b$".
So we see that we end up not with "$-b \geq a \geq b$" (which would mean "$-b \geq a$ and $a \geq b$") but rather with "$-b \geq a$ or $a \geq b$".
