Prove that $\sqrt{x} > \ln x$ for all $x>0$ with a study of function

Put $t = \sqrt{x} \implies e^t > t^ 2, 0 < t < 2$. Consider $f(t) = e^t - t^2 \implies f’(t) = e^t - 2t > 1+ t+ \frac{t^2}{2} - 2t = \frac{1}{2} + \frac{(t-1)^2}{2} > 0 \implies f(t) > f(0) = 1 > 0 \implies e^t > t^2$ .

If $x\in (0,1]$ then is obvius that $\sqrt{x}>\ln{x}$, Define $$f(x)=\frac{\sqrt{x}}{\ln{x}}$$ With $x\in(1,\infty)$, then $f(x)>0$ for all $x$, and has minimun in $x=e^2$, then: $$f(x)\geq\frac{e}{2}>1\implies\frac{\sqrt{x}}{\ln{x}}>1$$