Norm inequality (supper bound) Do you think this inequality is correct? I try to prove it, but I cannot. Please hep me.
Assume that $\|X\| < \|Y\|$, where $\|X\|, \|Y\|\in (0,1)$ and $\|Z\| \gg \|X\|,\|Z\| \gg \|Y||$.
prove that
$$\|X+Z\|-\|Y+Z\| \leq \|X\|-\|Y\|$$
and if $Z$ is increased, the left hand side become smaller.
I pick up some example and see that this inequality is correct but I cannot prove it.
Thank you very much.
 A: The inequality is false as stated. Let 
$$ \begin{align}
X &= (0.5,0)\\
Y &= (-0.7,0)\\
Z &= (z,0), 1 \ll z
\end{align}$$
This satisfies all the conditions given. We have that
$$ \|X + Z\| - \|Y + Z\| = z + 0.5 - (z - 0.7) = 1.2 \not\leq -0.2 = \|X\| - \|Y\| $$

From the Calculus point of view, in $n$ dimensions, we can write
$$ \|X\| = \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2} $$
We have that when $\|X\| \ll \|Z\|$, we can approximate $\|X+Z\| \approx \|Z\| + X\cdot \nabla(\|Z\|)$. Now, $\nabla(\|Z\|) = \frac{Z}{\|Z\|}$ by a direct computation, so we have that
$$ \|X + Z\| - \|Y + Z\| \approx (X-Y) \cdot \frac{Z}{\|Z\|} $$
From this formulation we see that even in the cases where $X,Y$ are infinitesimal the inequality you hoped for cannot hold true. However, 
the right hand side of this approximation can be controlled by Cauchy inequality to get (using that $Z / \|Z\|$ is a unit vector). 
$$ (X-Y) \cdot \frac{Z}{\|Z\|} \leq \|X - Y\| $$
So perhaps what you are thinking about is the following corollary of the triangle inequality
Claim: If $X,Y,Z$ are vectors in $\mathbb{R}^n$, then 
$$ \|X + Z\| - \|Y + Z\| \leq \|X - Y \| $$
Proof: We write 
$$ X + Z = (X - Y) + (Y + Z) $$
so by the triangle inequality
$$ \|X + Z\| = \|(X - Y) + (Y+Z)\| \leq \|X - Y\| + \|Y + Z\| $$
rearranging we get
$$ \|X + Z\| - \|Y + Z\| \leq \|X - Y\| $$
as desired. 
Remark: if we re-write the expression using $-Z$ instead of $Z$, the same claim is true in an arbitrary metric space: Let $(S,d)$ be a metric space. Let $x,y,z$ be elements of $S$. Then 
$$ d(x,z) - d(y,z) \leq d(x,y) $$.
