Prove that |a -b| $\geq$ ||a| - |b|| $\forall$ a,b $\in$ $\mathbb{R}$

Question

Here is my proof can anyone check if it needs more detail or if it's good?

Since a = a - b + b, then by the triangle inequality, |a| $\geq$ ||a| - |b|| + |b| so that |a| - |b| $\geq$ ||a| - |b|| since ||a| - |b|| = |a| - |b| $\therefore$ |a - b| $\geq$ ||a| - |b||

Answer 1

Since both sides are non-negative, the inequality holds if and only if $$(a-b)^2\ge (|a|-|b|)^2,$$ that is, if and only if $$a^2-2ab+b^2\ge a^2-2|ab|+b^2.$$ But this inequality is obvious.

Answer 2

If $a,b \in \mathbb{R}$, then $a = tb$ for some $t \in \mathbb{R}$. Then $|a-b| = |b||t - 1|$ and $||a| - |b|| = |b|||t|-1|$. If $t \geq 0$ then $|a-b| = ||a| - |b||$; if $t < 0$ then $||t| - 1| = |-t-1|$, so $|a-b| > ||a| - |b||$.