An elementary proof for a bound on $x \log x$ During one of our information theory classes, the Professor used the following bound to prove a result.
For any $x,y \in (0,1)$, $x \neq y$, show that 
$$|x \log(x) - y \log(y) | \leq |x-y|\log \left(\frac{1}{|x-y|}\right)$$
Note that RHS is positive when put this way. Also note that base of the logarithm is immaterial here(I take base $e$ to simplify calculations). Looking at $|x-y|$, I thought of using mean value theorem. W.L.O.G let $x > y$. Then by applying mean value theorem to $u \log u$, we have
$$x \log(x) - y \log(y) = (x-y)(1 + \log z) = (x-y)(\log ez)$$
for some $z \in (y,x)$. Now I do not know how to show (after removing logs).
$$ez \leq \left(\frac{1}{x-y}\right)$$
I'd like to know if my approach has any promise as well as any alternate ways to get this. As always any hints are appreciated.
Edit: Looks like I made an error while posting this problem, apparently the result also required that $|x-y| \le 0.5$. Kindly note this.
 A: Let we assume $x,y\in(0,1)$ and $|x-y|=h$. We want to prove:
$$ \sup_{\substack{x,y\in (0,1)\\ |x-y|=h}}\left|x \log x-y\log y\right|\leq -h\log h \tag{1}$$
but since $f(x)=x\log x$ is a convex function on $I=(0,1)$, it is enough to prove $(1)$ for $\min(x,y)\to 0^+$ and $\max(x,y)\to 1^-$. The first case is trivial, since $\lim_{x\to 0^+}x\log x = 0$.
So we just have to check:
$$ \forall h\in(0,1),\qquad -(1-h)\log(1-h) \leq -h\log h \tag{2}$$
but that holds only if $\color{red}{h\leq\frac{1}{2}}$.
A: From what I tried, the mean value approach will not work. So I found another approach using the method of secants. Given a function $f$, define for any two real numbers $b$ and $a$, $b \ne a$
$$R(b,a) = \frac{f(b)-f(a)}{b-a}$$
Now for a convex function (you need nothing more than the definition), we can show that for $a < b < c$
$$R(c,b)\ge R(c,a) \ge R(b,a)$$
We can further deduce for convex functions that if $a< b$, $c < d$, $a < c$ and $b<d$, 
$$R(b,a) \le R(d,c)$$
Flip all inequalities for concave functions. Now in my problem, let $f(u) = -u\log(u)$. Set $f(0)=0$ here. Note that this is a concave function. Now, assuming $x > y$ in the original question, let $x-y = \tau$. Thus we get by putting $a=0$, $b=\tau$, $c=y$ and $d=x=y+\tau$, that 
$$R(x,y) \leq R(\tau,0)\Rightarrow \frac{f(x)-f(y)}{x-y} \leq \frac{f(\tau)}{\tau} \\
\Rightarrow f(x) -f(y) \leq f(\tau)$$
Similarly noting that $f(1)=0$ and putting $a=y$, $b=x=y+\tau$, $c=1-\tau$, $d=1$, we get
$$f(x) -f(y) \geq -f(1-\tau)$$ 
Thus, noting that $f(u) \geq 0$ for $u\in(0,1]$, we get
$$|f(x) -f(y)| \leq \max\{f(\tau),f(1-\tau)\}$$
Since $\tau \leq 1/2$, $f(\tau) \geq f(1-\tau)$ which completes the proof. $\blacksquare$
