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.
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.