How to prove this inequality $ \left|\frac{{\sin x}}{x} - \frac{{\sin y}}{y}\right| \le \sqrt {2\left|\frac{1}{x} - \frac{1}{y}\right|}$ $$
\left|\frac{{\sin x}}{x} - \frac{{\sin y}}{y}\right| \leqslant \sqrt {2\left|\frac{1}{x} - \frac{1}{y}\right|}
$$
I tried to prove this by using mean value theorem, but I failed. Could you help me? Thanks!
 A: It is not true in general. For example, take
$x=\pi$, $y=\frac{\pi}{2}$
Then 
$|x\sin y - y\sin x| = \pi > \sqrt 2 |x - y|= \frac{\pi}{\sqrt 2}$
A: For $(x,y)\in[0,1]^2$, define $g(x,y)=x\sin y$. So $|\nabla g|=\sqrt{\sin^2y+x^2\cos^2y}\le1$
So by MVT $|g(x,y)-g(y,x)|\le 1\times\sqrt{(x-y)^2+(y-x)^2}=\sqrt{2}|x-y|$.
$$\therefore|x\sin y-y\sin x|\le\sqrt{2}|x-y|\;\forall\;x,y\in[0,1]$$
A: This only deals with $0\leq x<y\leq \pi$. Let $\phi(t)=\frac{\sin t}{t}$, so that
$\phi'(t)=\frac{t\cos t -\sin t}{t^2}=\big(\cos t -\frac{\sin t}{t}\big)\frac{1}{t}$. By Cauchy-Schwartz's inequality
$$\begin{align}
\Big|\int^y_x\big(\cos t -\frac{\sin t}{t}\big)\frac{1}{t}\,dt\Big|&\leq \sqrt{\int^y_x\big|\cos t -\frac{\sin t}{t}\big|^2\,dt}\sqrt{\int^y_x\frac{1}{t^2}\,dt}\\
&=\sqrt{\Big|\frac{1}{x}-\frac{1}{y}\Big|}\,\sqrt{\int^y_x\big|\cos t -\frac{\sin t}{t}\big|^2\,dt}
\end{align}$$
For $0\leq x<y\leq\pi$, we have
$$\begin{align}
\Big|\frac{\sin x}{x}-\frac{\sin y}{y}\Big|&\leq\sqrt{\Big|\frac{1}{x}-\frac{1}{y}\Big|}\sqrt{\int^\pi_0\big|\cos t -\frac{\sin t}{t}\big|^2\,dt}
\end{align}$$
A numerical estimation gives  $\int^\pi_0\big|\cos t -\frac{\sin t}{t}\big|^2\,dt\approx1.570796\ldots<1.6=\frac{8}{5}$. It follows that
$$\begin{align}
\Big|\frac{\sin x}{x}-\frac{\sin y}{y}\Big|&\leq \sqrt{\frac{8}{5}\Big|\frac{1}{x}-\frac{1}{y}\Big|}
\end{align}
$$
Outside that range the problem seems more difficult
A: For some $\psi$ we have
$$|x\sin y - y\sin x| = \sqrt{x^2+y^2}|\sin(x+\psi)|\le \sqrt{x^2+y^2}$$
And it is easy to see, that
$$x^2+y^2\le 2(x-y)^2$$
