Simple Normed Space Inequality

Question

Let $(V,\|\cdot\|)$ be a normed vector space. Let $x,y,x',y' \in V$. Say I want to estimate $$\left| \|x\|-\|x'\|-(\|y\| - \|y'\|) \right|.$$

Does the following chain of inequalities hold?:

\begin{eqnarray*} \left| \|x\|-\|x'\|-(\|y\| - \|y'\|) \right| &\leq& \left| \|x-x'\|-(\|y\| - \|y'\|) \right|\\ &\leq& \left| \|x-x'\|-(\|y-y'\|) \right| \\ &\leq& \|(x-x')-(y-y')\|) \end{eqnarray*}

For some reason I'm having a bit of a brain fart and can't seem to justify or disprove the 1st to 2nd inequality.

Answer 1

Consider $\mathbb{R}^2$ with the usual orthonormal basis $e_1, e_2$.

Take \begin{align*} x &= e_1 \\ x' &= e_2 \\ y &= 100e_2 \\ y' &= - e_2. \end{align*}

Answer 2

The overall inequality doesn't work, either. Again in $\mathbb{R}^2$, take $$ x = \tfrac{1}{\sqrt{2}}(-1, 1), \qquad x' = \tfrac{1}{\sqrt{2}}(1, 1), \qquad y = (-\sqrt{2}, 0), \qquad y' = (0, 0). $$ Then $$ \left| \|x\| - \|x'\| - (\|y\| - \|y'\|) \right| = \left| 0 - \sqrt{2} \right| = \sqrt{2} $$ but $$ \left\|(x-x') - (y-y') \right\| = \| (-\sqrt{2}, 0) - (-\sqrt{2}, 0) \| = 0, $$ so your desired estimate is not valid.