Sharper than Mean Value Inequality Prove that 
$$|x\ln x-y\ln y| \le |x-y|^{1-1/e}$$
for $0<y<x\le 1$
Using the Mean Value theorem, all what I found that there exist $c\in (y,x)$ such that 
$$|x\ln x-y\ln y| \le |x-y|\max_{c\in (y,x)}|1+ln(c)|$$
 A: $\max_{c\in (y,x)}|1+\ln c| \leqslant |1 + \ln x|$.
When $e^{-1}<x\leqslant 1$, $1 + \ln x > 0$, let $f(x) = x^{\frac1e}(1 + \ln x)$, its derivative
$$\begin{align}f'(x) &= \frac1e x^{\frac1e - 1}(1 + \ln x) + x^{\frac1e} \frac1x \\
&= x^{\frac1e - 1}\left( 1 + \frac{1 + \ln x}{e}\right) \\
&> 0
\end{align}$$
Thus $\forall e^{-1}<x\leqslant 1, \,\, f(x) \leqslant f(1) = 1$.
When $0 < x \leqslant e^{-1}$, let $f(x) = -x^{-\frac1e}(1 + \ln x)$. Its derivative
$$\begin{align}f'(x) &= -\frac1e x^{\frac1e - 1}(1 + \ln x) - x^{\frac1e} \frac1x \\
&= -x^{\frac1e - 1}\left( 1 + \frac{1 + \ln x}{e}\right) 
\end{align}$$
$$ f'(x) = 0 \,\, \mbox{ iff. } \,\, x = x_0 =  e^{-1-e}.$$ 
The value of $f$ at $x_0$ is :
$$ f'(x_0) = -e^{\frac{-1-e}{e}} \cdot (1-1-e) = e^{1 - \frac1e - 1} = e^{-\frac1e} < 1$$
Notice that $ f'(x) < 0$ when $x_0 < x \leqslant e^{-1}$ and $ f'(x) > 0$ when $ x < x_0$. $f(x)$ achieves its maximum at $x_0$.
So we conclude that $x^{\frac1e}|1 + \ln x|$ is less than 1 when $x \in (0,1]$.
Then we have 
$$\begin{align}|x - y|^{\frac1e}|x \ln x - y \ln y| &\leqslant x^{\frac1e} |x - y|\max_{c\in (y,x)}|1+\ln c|  \\
&\leqslant |x - y| x^{\frac1e} |1 + \ln x|\\
&\leqslant |x - y|.
\end{align}$$
Thus $|x \ln x - y \ln y| \leqslant |x - y|^{1 - \frac1e}$.
