I stumbled upon the following inequality:
$$\Vert x+hz-(x+y)-(p-(x+y))\Vert_2 \geq \Vert p-(x+y)\Vert_2-\Vert x+hz-(x+y)\Vert_2$$
where $p,x,y,z \in \mathbb{R}^n$.
My question is: Is this some kind of simple triangle inequality?
Because to me, it totally looks like that but I can't figure it out, any approach ended up with something like a $+$ instead of a $-$ or a $\leq$ instead of a $\geq$. I need to know if the author is using a rather simple trick with the norms (aka. triangle inequality) or if there's more about it.
I (once again) feel like I'm missing out on something completely obvious, any help is appreciated.
Yes, it is a second form of triangle inequality :
$|a-b| \geq ||a|-|b||$ and in particular $|a-b| \geq |a|-|b|$
On simple example (let's say we are in the 2-dimensionnal euclidian space), you can understand it as follows : when substracting two vectors, you can see that the "smallest" vector you can achieve is by taking $a$ and $b$ positively related ($a=\lambda b,\lambda \geq 0$), in which case the norm $||a-b||$ is exactly $||a||-||b||$ (or $||b||-||a||$ if $\lambda \leq 1$). Otherwise, it is greater than that.
Proof : let $|.|$ be a norm, $u,v$ two elements, then
$|u|=|u-v+v| \leq |u-v| + |v|$ hence $|u| - |v| \leq |u-v|$. You can do the same swapping $u$ and $v$ to get the other way around, and finally get the inequality for the absolute value.