Where the $\log$ is the natural log of $x$. It is only required for the interval $(0,1)$.
I tried square both side (since both of them are positive) $ x^2\log^2{(x)}\le x$ , which implies $ x\log^2{(x)}\le 1$. However, I have no clue how to proceed from here. I also tried solve it in exponential form, which would be $e^{(- x\log{(x)})} = x^{-x} \le e^{\sqrt{x}}$, but don't know how to move on from here as well...
In addition, I do think differentiating $x\log{(x)}+\sqrt{x}$ would help at all since even if it can be proved that it is an increasing function for all $x>0$, I still cannot conclude that $- x\log{(x)}$ is strictly less than $\sqrt{x}$ since the domain of $x\log{(x)}+\sqrt{x}$ is an open interval