Proving $\sup(|f|) - \inf(|f|) \leq \sup(f) - \inf(f)$ I appreciate if you could give me some hints on how to prove that :

$$\sup(|f|) - \inf(|f|) \le \sup(f) - \inf(f)$$
Here is a strategy. Since you asked for hints, I let you check the details.

This is not really a fact about functions. It is rather a property of subsets $S$ of $\mathbb{R}$. So take such a set $S$ and denote $|S|:=\{|s|\;;\;s\in S\}$. We want
$$
\sup |S|-\inf|S|\leq \sup S-\inf S.
$$

We have an alternative.

Case 1: $\sup |S|=\sup S$. Clearly $\inf S\leq \inf |S|$. So the inequality holds.

Case 2: $\sup|S|=-\inf S$. Check that $-\inf|S|\leq \sup S$. The inequality follows.
Let $c = \pm 1$,then
$sup(f) - inf(f) \leq \left| sup(f)-inf(f) \right| \leq c*[sup(f)-inf(f)] = c*sup(f)-c*inf(f) = sup(c*f)-inf(c*f) \leq  sup(\left| f \right|) - inf(\left| f \right|) $