Showing $f(z) \le 2 \sqrt{ \varepsilon} f(y)$ if $z < \varepsilon y$ for given $f$ Is is possible to show $f(z) \le 2 \sqrt{ \varepsilon} f(y)$ if $z < \varepsilon y$ for $f(x) = \sqrt{|x|\log \frac{1}{|x|}}$?
I would like to use this inequality in a proof but I don't know wether it is true and how I can show it.
Any hints would be great!
 A: First of all, for $0<z<1$ we have $\log{z}<0$ which is a problems for defining $\sqrt{\log{z}}$. As a result $z\geq 1$ or, in terms of $f(x)$, we should have $\log{\frac{1}{|x|}}\geq 0 \Rightarrow \frac{1}{|x|}\geq 1 \Rightarrow 0< |x|\leq 1$. This is the domain of $f(x)$.
Obviously $f(x) \geq 0$,  $f(1)=f(-1)=0$. Also $f(x)=f(-x)$ as a result we can restrict the study to $0< x\leq 1$ (the other side is symmetric).
Function $f(x)$ has a maximum at $x=\frac{1}{e}$ and is ascending on $\left(0,\frac{1}{e}\right)$. This means that for 
$$0<z<y<\varepsilon y \leq \frac{1}{e} \Rightarrow f(z) \leq f(y) < 2\sqrt{\varepsilon}f(y)$$
However, if we still assume the inequality is true for the entire $0< x\leq 1$ :


*

*For $y=1$ and $0<z<\varepsilon y=\varepsilon \leq 1$, this leads to $0\leq f(z)\leq 2\sqrt{\varepsilon}f(1)=0$ which is only possible for $z=1$ - contradiction.

*For $y=1-\frac{1}{n},n \in \mathbb{N}\setminus \{0,1\}$ and $z=\frac{1}{e}$ we can tune $0<\frac{1}{e}<\varepsilon \left(1-\frac{1}{n}\right)<1$, for example $\varepsilon=\frac{1}{2}$. Using this inequality $$f\left(\frac{1}{e}\right)=\frac{1}{\sqrt{e}}\le 2\sqrt{\varepsilon}f\left(y\right)=2\sqrt{\varepsilon}\sqrt{y\log{\frac{1}{y}}} \le 2\sqrt{\varepsilon}\sqrt{y\left(\frac{1}{y}-1\right)}=2\sqrt{\varepsilon}\sqrt{1-y}=2\sqrt{\frac{\varepsilon}{n}}$$ or simply $$\frac{1}{\sqrt{e}} \le 2\sqrt{\frac{\varepsilon}{n}}$$ which is wrong for large $n$ - another contradiction. In fact, an infinite sequence of contradictions if we note $y_n=1-\frac{1}{n}$.


So, there is no definitive answer as long as we keep the entire domain $0< |x|\leq 1$, further restrictions are required.
