TITLE: How to prove this inequality $\left|x\sin{\frac{1}{x}}-y\sin{\frac{1}{y}}\right|<2\sqrt{|x-y|}$?
QUESTION [17 upvotes]: For any real numbers $x,y\neq 0$,show that
$$\left|x\sin{\dfrac{1}{x}}-y\sin{\dfrac{1}{y}}\right|<2\sqrt{|x-y|}$$
I found this problem when I dealt with this problem. But I can't prove it. Maybe the constant $2$ on the right hand side can be replaced by the better constant $\sqrt{2}$?
Thank you.

REPLY [2 votes]: This is a partial alternative solution for the case $\left|\frac{1}{x}-\frac{1}{y}\right|\leq 1$.
We can clearly assume that both $x$ and $y$ are positive, then, through the substitutions $x\to 1/x,y\to 1/y$, prove that:
$$ \forall x,y>0,x\neq y,\quad (y\sin x-x\sin y)^2 < 4xy|y-x|. $$
The LHS can be written as:
$$ \left((y-x)\sin x + x\,(\sin x-\sin y)\right)^2, $$
that, by the Cauchy-Schwarz inequality, satisfies:
$$ \left((y-x)\sin x + x\,(\sin x-\sin y)\right)^2 \leq (x^2+\sin^2 x)\left((y-x)^2+(\sin y-\sin x)^2\right), $$
and the RHS is less than $4x^2(y-x)^2$, since $\sin x$ is a $1$-Lipschitz function. 
By exchanging $x$ and $y$, we have:
$$ (y\sin x-x\sin y)^2 < \min\left(4x^2(y-x)^2,4y^2(y-x)^2\right)<4xy(y-x)^2,$$
so the inequality is clearly true if $|y-x|\leq 1$.