Let $f(x)=\sqrt{x} - \ln{x}$. Then it is sufficient to show that the minimal value of $f(x)$ is positive for $x > 0$.

$$f'(x)=\frac{1}{2\sqrt{x}}-\frac{1}{x}=\frac{\sqrt{x} - 2}{2x}$$

Set $f'(x)=0$ gives $x=4\implies f(4)=2-\ln{4} > 0$. Meanwhile $f'(x) < 0$ for all $0 < x < 4$ and $f'(x) > 0$ for all $x > 4$. This means that $f(x)$ is decreasing in $(0, 4)$ and increasing in $(4, \infty)$. In other words $f(4)$ is the global minimal which is a positive number.