Inequality Related to $\arctan$ function How can I prove that for any $x$ and $y$ in $\mathbb{R}^n$, 
$$
\left|\frac{2}{\pi}\arctan(|x|)\frac{x}{|x|}- \frac{2}{\pi}\arctan(|y|)\frac{y}{|y|}\right| < M|x-y|$$ for some $M > 0$.
I tried it using the fact that derivative of $\arctan$ function is always less than equal to $1$.
 A: We have $\frac{x}{|x|} \arctan |x| = \arctan x$ for $x\in\mathbb{R}$.
Furthermore, $\arctan' x = \frac{1}{1+x^2} \leq 1$.
Starting from $0 < \frac{2}{\pi} \frac{1}{1+x^2} \leq \frac{2}{\pi}$, you get your inequality by integrating from $x$ to $y$.
(and in fact $M= \frac{2}{\pi}$)
A: Denote 
$$
f(x)=\frac{2}{\pi}\frac{\arctan|x|}{|x|} x
$$
then by mean value theorem for vector valued functions we have 
$$
|f(x)-f(y)|\leq \left(\sup\limits_{z\in[x,y]}\Vert(Df)(z)\Vert\right)|x-y|
$$
where $Df$ is a Jacobi matrix of $f$. It is remains to show that this supremum is always finite and independent of choice of $x,y\in\mathbb{R}^n$. 
Note that
$$
(Df)(z)_{ij}=\frac{\partial}{\partial z_j}\left(\frac{2}{\pi}\frac{\arctan|z|}{|z|} z_i\right)=
\left(\frac{1}{1+|z|^2}-\frac{\arctan|z|}{|z|}\right)\frac{z_i z_j}{|z|^2}+\frac{\arctan|z|}{|z|}\delta_{i,j}
$$
This value is bounded by some constant $C_{i,j}$ independent of $z$ because functions 
$$
\frac{1}{1+|z|^2}\qquad\frac{\arctan|z|}{|z|}\qquad\frac{z_i z_j}{|z|^2}
$$
are bounded. Thus all elements of matrix $(Df)(z)$ are bounded by some constant $C=\max C_{i,j}$, and this constant is independent of vector $z$. Hence operator norm $\Vert (Df)(z)\Vert$ of matrix $(Df)(z)$ is also bounded by another constant $M$ independent of vector $z$. Finally,
$$
|f(x)-f(y)|\leq \left(\sup\limits_{z\in[x,y]}\Vert(Df)(z)\Vert\right)|x-y|\leq M|x-y|
$$
A: The following argument does not use multivariate calculus:
For $r\geq0$ put $r':=\arctan r$. Looking at the graph of $\arctan$ one immediately sees that $r'\leq r$ and $|r'-s'|\leq|r-s|$.
It suffices to consider the two-dimensional situation. Given the two points $$x=(r,0)\ ,\quad y=s(\cos\theta,\sin\theta)$$
with $r\geq0$, $s\geq0$ we consider the "scaled" points $$x':=(r',0)\ ,\qquad y':=s'(\cos\theta,\sin\theta)\ .$$ In terms of $x'$, $y'$ your "new distance" $\delta(x,y)$ satisfies
$$\eqalign{\Bigl({\pi\over2}\delta(x,y)\Bigr)^2=|x'-y'|^2&=r'^2+s'^2-2r's'\cos\theta\cr &=(r'-s')^2+2r's'(1-\cos\theta)\cr &\leq(r-s)^2+2rs(1-\cos\theta)\cr &=|x-y|^2\ .\cr}$$
This proves
$$\delta(x,y)\leq{2\over\pi}|x-y|\ .$$
Letting $x=0$, $y\to0$ one easily verifies that this inequality is best possible.
