prove that for all $a,b \in \mathbb{R}$, $ | |a|-|b|| ≤ |a-b|$
so what I have thought of so fat is I tried to use the definition of absolute value then getting $b-a ≤ |a|-|b| ≤ a-b.$ from there I'm getting $2b≤|a|-|b|≤2a$. Here is where I encountered the problem of where to proceed. How could I use the fact to keep going to proof the statement
It is just a matter of conveniently applying the triangle inequality twice. Since it is true that $|x - y| = |y - x|$, the answer follows \begin{align*} &\begin{cases} |x| = |x - y + y| \leq |x - y| + |y|\\ |y| = |y - x + x| \leq |y - x| + |x| \end{cases} \Longrightarrow \begin{cases} |x - y| \geq |x| - |y|\\ |x - y| \geq |y| - |x| \end{cases}\Longrightarrow\\\\ & |x - y| \geq \max\{|x| - |y|,-(|x|-|y|)\} \Longrightarrow |x - y| \geq ||x| - |y|| \end{align*}
Since $|x| = \mathrm{max} \{ x, -x \}$ for all $x$, it suffices to prove that $|a| - |b| \le |a-b|$ and $|b| - |a| \le |a-b|$.
The triangle inequality yields $$|a| = |(a-b) + b| \le |a-b| + |b| \quad \text{and} \quad |b| = |(b-a) + a| \le |b-a| + |a|$$ The desired inequalities above follow by subtracting the appropriate term from each side (and using $|a-b| = |b-a|$).
