Hard Euclidean norm inequality $\big\|\frac{x}{\|x\|}-\frac{y}{\|y\|}\big\|\leq\frac{2}{\|x\|+\|y\|}\|x-y\|.$ So, today I had a Calc $3$ exam (it already finished) with a bonus question, which I couldn't (and still can't) answer. It's the following:

Let $\|\cdot\|:\mathbb{R}^n\to\mathbb{R}$ be the Euclidean norm. Prove that for any two vectors $x,y\in\mathbb{R}^n$ with $x\neq 0\neq y$ the following inequality holds: $$\left|\left|\frac{x}{\|x\|}-\frac{y}{\|y\|}\right|\right|\leq\frac{2}{\|x\|+\|y\|}\|x-y\|.$$

I'm very curious as to how I can attack this question. Any hints as well as a full answer is appreciated. I've tried using the triangle inequality (which is, basically, all I could think about) with no luck. I guess I could try using the definition of the euclidean norm letting $x=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_n)$ and try something, but I don't want to go down that path (and I don't think it would lead me anywhere, either).
 A: Let $x = r u$ and $y = s v$ where $u$ and $v$ are unit vectors, $\|x\| = r$ and $\|y\| = s$.  Then your inequality says
$$\|u - v\| \le 2 \frac{ \|r u - s v\|}{r+s} \ \text{for}\ r,s>0$$
By homogeneity,  we may assume $r+s=1$.  Thus the inequality becomes
$$ \|u - v\| \le 2 \|r u - (1-r) v\| \ \text{for}\ 0 < r < 1$$
The right side is convex as a function of $r$.  Now there is a linear isometry
that interchanges $u$ and $v$, so $\|r u - (1-r) v\| = \|(1-r) u - r v\|$.
Thus the minimum must occur at $r = 1/2$.
A: Let $x=ur$ and $y=sv$ where $r,s\in \Bbb R^+$ and $\|u\|=\|v\|=1.$
Let $c=s/r.$ Then $c>0$.
The LHS of the inequality is $\|u-v\|.$ The RHS is $\frac {2}{1+c} \|u-cv\|.$
We have $\|u\|^2=\|v\|^2=1.$ And $1+c>0$. So we have $$\|u-v\|\,\le \,\frac {2}{1+c} \|u-cv\| \iff$$ $$ (1+c)\cdot \|u-v\|\,\le\, 2 \|u-cv\| \iff$$ $$ (1+c)^2\cdot \|u-v\|^2\,\le\, 4\|u-cv\|^2 \iff $$ $$(1+c)^2\cdot (\|u\|^2+\|v\|^2-2u\cdot v)\,\le \,4\cdot (\,\|u\|^2+c^2\|v\|^2-2c(u\cdot v)\,) \iff$$ $$ (1+c)^2\cdot (2-2u\cdot v)\le 4(1+c^2-2c(u\cdot v)) \iff $$ $$(-2u\cdot v)\cdot (\,(1+c)^2-4c)\,\le\, 4 (1+c^2)-2 (1+c)^2 \iff$$ $$ (-2u\cdot v)(1-c)^2\le 2(1-c)^2.$$ Now $|-2u\cdot v|\le 2\|u\|\cdot \|v\|=2.$ So the absolute value of the LHS of the last line above is at most $2(1-c)^2.$
