$\left|\frac{x}{|x|}-\frac{y}{|y|}\right|\leq |x-y|$, for $|x|, |y|\geq 1$? 
Let $(V,|\cdot|)$ be a normed space. Let $x, y\in V.$ If $|x|,|y|\geq 1$, then does it follow that
  $$\left|\frac{x}{|x|}-\frac{y}{|y|}\right|\leq |x-y|\;?$$

Actually I got a hint in form of a picture, but I failed to use it.

 A: It is false in general. Take e.g. $x=(1 \ \  1)$, $y=(\frac{4}{3}  \ \ \frac{1}{3})$ in the sup norm. Then 
$$ \left|\frac{x}{|x|}-\frac{y}{|y|}\right|= \left|(1 \ \ 1) - ( 1\ \ \frac14)\right| =\frac34 >\left|{x}-{y}\right| =\frac23$$.
On the other hand the following weaker inequality is always true for $|x|,|y|\geq 1$:
$$ \left|\frac{x}{|x|}-\frac{y}{|y|}\right|=
\left|\frac{x|y| -y|y| + y|y| - y|x|}{|x||y|}\right|
\leq \frac{\left|x-y\right|+ \left|\,|x|-|y|\,\right|}{|x|}  \leq 2 |x -y|$$
A: NO. Let $V=\mathbb R^2.$ Let $\|(u,v)\|=\max (|u|,|v|).$ Let $x=(1,1)$ and $y=(1/2,3/2).$ The norm of $x/\|x\|-y/\|y\|$ is $2/3.$ But $\|x-y\|=1/2.$
A: A trail:
As there are only two vectors involved, we may suppose we're in a normed space of dimension $2$. 
I conjecture that we may as well suppose we're in $\mathbf C$, and the norm is the modulus, up to an isometry.
In this case, we may as well suppose $\lvert x\rvert=1$, and even that $x=1$. So let $y=r\mathrm e^{i\theta}\enspace(r\ge 1)$. We have to show
$$\lvert r\mathrm e^{i\theta}-1\rvert\ge \lvert\mathrm e^{i\theta}-1 \rvert,$$
i.e.
\begin{align}
(r\mathrm e^{i\theta}-1)(r\mathrm e^{i\theta}+1)&\ge(\mathrm e^{i\theta}-1)(\mathrm e^{i\theta}+1) \\\iff r^2-2r\cos \theta+1&\ge2(1-\cos\theta).
\end{align}
Now, as a function of $r$, $r^2-2r\cos \theta+1$ is increasing for $r\ge 1$ and $\theta\not\equiv 0\mod 2\pi$, so $$ r^2-2r\cos \theta+1\ge 1-2\cos \theta+1=2(1-\cos\theta).$$
