EDIT: Counter-example found. Statement is FALSE.
However, I think the argument still has value. It is true if you restrict the domain to $[0.223,0.716]$. Maybe $[\frac{3}{10},\frac{7}{10}]$ so it’s less “obvious”?
Prove inequality $|y \ln{y} - x \ln{x}| < 2 |\ln{\frac{1}{|y-x|}}|$ , when $x,y \in (0,1]$, $x \neq y$.
Assume WLOG $y > x$.
I have tried Mean Value Theorem. For some $c \in (0,1)$: \begin{align*} |y \ln{y} - x \ln{x}| &= (1+\ln{c})(y-x) = (\ln(e)+\ln(c))(y-x)\\ & \leq 2 \ln{\frac{1}{(y-x)}} \text{ if and only if } \frac{(y-x)^2}{e} \leq c \leq \frac{1}{e(y-x)^2} \end{align*}
So the inequality is true when $$ \frac{(y-x)^2}{e} \leq x \leq y \leq \frac{1}{e(y-x)^2} $$
in which $c$ is between $x$ and $y$. Now we need to prove for the following cases in a different way: \begin{align*} x < \frac{(y-x)^2}{e} &\implies x^2 - (2+e)x + 1 > 0 : \text{when 0<x<0.222ish}\\ \frac{1}{e(y-x)^2} < y &\implies y > e^{-1/3} : \text{when y > e^-1/3 = 0.716ish } \end{align*}
as $1-x>y-x>y$. I think this statement should be true because some of the cases can be taken care by MVT, and cases where it can’t be solved by MVT are when $x$ and $y$ are near the endpoints.
I tried using taylor $\ln(1+x)$ but I can’t get anything. Any help would be appreciated.
It's wrong. Try $x=1$ and $y=0.05.$
In this case $RHS-LHS=-0.047...<0.$