Is the inequality $ |f(x)| > g(x) $ equivalent to $ f(x) > g(x) $ OR $ - f(x) > g(x) $ in the set of all real numbers?
My answer is yes.
I tried. First, I proof $$|f(x)| <g(x) \Leftrightarrow -g(x) < f(x) < g(x). $$
- If $ g(x) \leqslant 0 $, then the inequality $ |f(x)| <g(x) $ has no solution and $ -g(x) < f(x) < g(x) $ has no solution too.
- If $ g(x) > 0 $, from $ |f(x)| <g(x) $, we have $ f^2(x) < g^2(x) $, and then $ -g(x) < f(x) < g(x) $.
Similarly, we have $$|f(x)| \leqslant g(x) \Leftrightarrow -g(x) \leqslant f(x) \leqslant g(x). $$
Negate the sentence $ |f(x)| \leqslant g(x) $ is $ |f(x)| > g(x) $, therefore $ |f(x)| > g(x) $ is equivalent to $ f(x) > g(x) $ or $ - f(x) > g(x) $.
Am I true? Is there another proof?
I can't get the point of such a long proof: $$|f(x)|=\begin{cases} f(x) &\text{if $f(x)\ge 0$}\\ -f(x) &\text{otherwise}\end{cases}$$
So $|f(x)|>g(x)$ means $f(x)>g(x)$ if $f(x)\ge 0$, $-f(x)>g(x)$ otherwise.
On the other hand, if the question was about the mutually exclusion, it's clearly false: let $f(x)=1$ and $g(x)=-2$, so both |f(x)| and $-f(x)$ are greater than $g(x)$ for all $x\in\mathbb{R}$.