Vector norm Inequality proof Does anyone know how to start proving this inequality
$$
\left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\| \leq \frac{4 \|x-y\|}{\|x\|+ \|y\|}
$$
The norm is a random norm on a vector space $V$
 A: Here is a somewhat complicated proof, I do not know if there is a simpler
and more natural one.
Let $a=||x||,b=||y||,c=||x-y||$. The inequality is easy if $a=b$, so
assume that it is not the case. Since the problem is symmetric in
$x$ and $y$, we can assume $a < b$.
Then $2a < 2b$, so $a+b < 3b-a$, $ \frac{a+b}{3b-a}<1$, and hence
$\frac{b^2-a^2}{3b-a} < b-a$.
The triangle inequality shows that $||y|| \leq ||y-x||+||x||$ ; we deduce
$c \geq  b-a > \frac{b^2-a^2}{3b-a}$. It follows that
$(3b-a)c > b^2-a^2$, $ 4bc > b^2-a^2+(a+b)c$ and hence
$\frac{4c}{a+b} > \frac{b-a+c}{b} $.
So it will suffice to show the following :
$$
\left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\| \leq \frac{b-a+c}{b}
\tag{1}
$$
To show (1), we use the identity
$$
\bigg(\frac{x}{a}-\frac{y}{b}\bigg)=
\frac{c}{b}\bigg(\frac{x-y}{c}\bigg)+
\frac{b-a}{b}\bigg(\frac{x}{a}\bigg)
 \tag{2}
$$
It follows from (2) that 
$$
\bigg(\frac{x}{||x||}-\frac{y}{||y||}\bigg)=
\frac{c}{b}\bigg(\frac{x-y}{||x-y||}\bigg)+
\frac{b-a}{b}\bigg(\frac{x}{||x||}\bigg)
 \tag{3}
$$
Applying the triangle inequality to (3), we see that
$$
 \left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\|
\leq \frac{c}{b}+\frac{b-a}{b} \tag{4}
$$
which concludes the proof.
A: To make life easier, first assume $\|x\| = \|y\| = 1$ which would reduce your inequality to a much more basic
$$
\|x-y\| \leq 2 \|x-y\|
$$
And now to prove the big result, you have it on the unit-scaled versions of the vectors. Can you take it from here?
