Inequality $\left|x\sin\frac{1}{x}-y\sin\frac{1}{y}\right|\leq\sqrt{2|x-y|}$

$\forall x,y>0$, then $$|x\sin\frac{1}{x}-y\sin\frac{1}{y}|\leq\sqrt 2\sqrt{|x-y|}$$

Is $\sqrt 2$ the minimum positive real number such that the inequality holds ?

I try to apply Mean value theorem, but no success, $f'$ is unbounded near zero

A related problem can be found in American Mathematical Monthly 6(June-July 1941), pp. 413-414

• – Nick Strehlke Jan 21 '14 at 21:15
• By considering the inverses of x and y, you will get rid of the singular behavior and obtain the well known sinc function. – Yves Daoust Jan 21 '14 at 21:26

First, I'd make the substitution $x\to1/x$, $y\to1/y$. The inequality becomes: $$\frac{\sin (x)}{x}-\frac{\sin (y)}{y}\leq \sqrt{2 \left(\frac{1}{x}-\frac{1}{y}\right)}$$ where I've dropped the absolute value for readability. Now consider the shape of the function $\frac{\sin (x)}{x}$. It's local extrema are progressively smaller; let's denote them with $x_i$. Hence if we have $y-x$ which is "big" we can always find a $z$ such that both $x$ and $z$ belong to the same $[x_i,x_{i+1}]$ interval and $f[z]=f[y]$. Effectively this means that the LHS of the inequality stays the same but the RHS decreases.
This shows that it is enough to prove the inequality when both $x$ and $y$ belong to the same $[x_i,x_{i+1}]$ interval - meaning they are close to each other.
In this case, I guess, you can successfully use the Taylor expansion of $\frac{\sin (x)}{x}$ around $2k\pi$ with 4 terms and prove the inequality.