Vector Inequality with ratio upper bound "Let $0\neq x\neq y$ be vectors in $\mathbb{R}^n$. Show that
$$\left|\frac{x-y}{|x-y|}-\frac{x}{|x|}\right|\leq2\frac{|y|}{|x|}"$$
I can see that if it holds for $(x,y)$ then it also holds for $(rx,ry)$ for $r\neq 0$. Thus, if $|y|\geq |x|$, we can take $|x|=1$ and then it is immediate by the triangle inequality. I don't see how to treat the case where $|y|<|x|$; any ideas?
 A: Write the question as follows:
$$
\left|\frac{x-y}{|x-y|}-\frac{x}{|x|}\right|\leq2\frac{|(x-y) - x|}{|x|}
$$
Now using for any vector $z$ that $z^2 = |z|^2$, we can square and get the equivalent
 $$
2 - 2 \cos(x-y,x)\leq 4 \Big( \frac{|x-y|^2}{|x|^2} + 1 -2 \frac{|x-y|}{|x|}\cos(x-y,x) \Big) 
$$
or
$$
0\leq 2\frac{|x-y|^2}{|x|^2} + 1+ \Big( 1 - 4\frac{|x-y|}{|x|} \Big) \cos(x-y,x)  
$$
Since $1 \ge \cos(x-y,x) \ge -1$ it suffices to show that 
$$
0\leq 2\frac{|x-y|^2}{|x|^2} + 1- \Big| 1 - 4\frac{|x-y|}{|x|} \Big|
$$
Letting $w = \frac{|x-y|}{|x|} \ge 0$ we consider two cases:


*

*$0.25 \ge w \ge 0$:
$$
0\leq 2w^2 + 1- \Big( 1 - 4w \Big) = 2w^2 + 4w
$$
which is evident.

*$ w \ge 0.25$:
$$
0\leq 2w^2 + 1 +\Big( 1 - 4w \Big) = 2w^2 +2 -  4w = 2 (w-1)^2
$$
which is also evident. $\qquad \square$
Remark: we see that the inequality is tight for $w=1$ (i.e. $|x-y| = |x|$) and $1 =| \cos(x-y,x)|$, hence for $y =0$ and any $x$. 
Also, the inequality is tight for $w=0$ (i.e. $|x-y| = 0$) and $1 =| \cos(x-y,x)|$, hence for  any $x$ and $y = ax$ with $a > 1$ and $a \to 1$. This case is not as obvious as the first one. Some explanation: Since $a >1$, $\frac{x-y}{|x-y|} = x \frac{1-a}{|1-a||x|} = -\frac{x}{|x|}$. Inserting this in the original inequality shows that it is tight:
$$
\left|\frac{x-y}{|x-y|}-\frac{x}{|x|}\right| = \left|-2 \frac{x}{|x|}\right| = 2 \leq 2\frac{|y|}{|x|} = 2a \frac{|x|}{|x|} {{{\tiny {\{a \to 1 \}}}\atop{\to}}\atop{{\tiny{\;}}}} 2
$$
Obviously $a >1$ is needed. 
There can be no other cases where the inequality is tight, as  $1 =| \cos(x-y,x)|$ is always required.
A: In Prove $\max \{|x|,|y|\}\cdot | x/|x| - y/|y| | \leq 2|x-y|$ an estimate was shown which holds for vectors as well:

Let $x, y \in \Bbb R^n \setminus \{ 0 \}$. Then
  $$
\max (\|x\|,\|y\|) \left\| \frac{x}{\|x\|} - \frac{y}{\|y\|} \right\| \leq 2\|x-y\| \, .
$$

The proof is identical to the proof here, only with the absolute value replaced by the norm.
Applied to our problem, we have
$$
\left\|\frac{x-y}{\|x-y\|}-\frac{x}{\|x\|}\right\| \le 2 \frac{\| y \|}{\max (\|x-y\|,\|x\|)}  \le 2 \frac{\| y \|}{\|x\|} \,.
$$
