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I'm looking to prove the following inequality

\begin{align} ||\frac{u}{||u||}-\frac{v}{||v||}|| \leq 2||u-v|| \end{align}

where $u$ and $v$ are elements of a Banach space such that $||u||$ and $||v||$ are greater than $1$.

I know that $||\frac{u}{||u||}||=1$ and have also tried using the triangle inequality.

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  • $\begingroup$ HAve you tried with the inverse triangle inequality? $\endgroup$ – Tito Eliatron Apr 11 at 18:43
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    $\begingroup$ @TitoEliatron Thanks for the hint. The reverse triangle inequality implies that $2|| ||u||-||v|| || < 2||u-v||$, which looks promising. Not sure where to go from there. $\endgroup$ – Bartolo Colon Apr 11 at 19:05
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Assume w.l.o.g that $\|y\|\leq \|x\|$. For convenience set $\|\frac{x}{\|x\|}-\frac{y}{\|y\|}\|=[x,y]$.

First consider the case when $\|y\|\leq (1-\frac{1}{2}[x,y])\|x\|.$ Then $$\|x\|\leq\|x-y\|+\|y\|\leq\|x-y\|+(1-\frac{1}{2}[x,y])\|x\|,$$ implying, along with our assumption, that $$[x,y]\leq2\|x-y\|/\|x\|\leq2\|x-y\|.$$ If $\|y\|\geq (1-\frac{1}{2}[x,y])\|x\|$, we have: \begin{align*} [x,y]\|x\|&=\|x-\frac{\|x\|y}{\|y\|}\|\\ & =||x-y+\frac{y}{\|y\|}(\|y\|-\|x\|)\|\\ &\leq \|x-y\|+\|y(\frac{\|x\|}{\|y\|}-1)\|\\ &=\|x-y\|+\|x\|-\|y\|\\ &\leq \|x-y\|+\|x\|+(\frac{1}{2}[x,y]-1)\|x\|. \end{align*} Rearranging this inequality and using the fact that $\|x\|\geq 1$ completes the proof.

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