Proof of an inequality including logarithms

Question

$0 <x,y \le 1$

$|y \log y-x\log x|\le|x-y|^{(1-\frac{1}{e})}$

I want to know how to solve this inequality.

Answer 1

$$| \log y^y-\log x^x|\le|x-y|^{(1-\frac{1}{e})}$$

Note that both sides of the inequality symmetric around the line $x=y$ on which the equality holds. Thus we can restric the discussion to the half plane $x>y$ where $$| \log y^y-\log x^x|=\left| \log \frac{x^x}{y^y}\right|$$ but for $x>y$ $$\log\frac{x^x}{y^y} =\log \left(\frac{x}{y^{y/x}}\right)^x=x \log \frac{x}{y^{y/x}}\ge x \log \frac{x}{y}>0$$ Therefore for $x>y$ the question is about $$ \log x^x-\log y^y\le(x-y)^{(1-\frac{1}{e})}$$

From here I would try to show that there is no point in the range where both partial derivative of $$f(x,y)= \log x^x-\log y^y-(x-y)^{(1-\frac{1}{e})}$$ vanishes and therefore the inequality holds always. But it could be tricky to do it.