Prove that $\sqrt{x} > \ln x$ for all $x>0$ with a study of function Prove that for all $x>0$ it is $\sqrt{x} > \ln x$.
I've defined the function $D(x)=\sqrt{x}-\ln x$, I've studied its first derivative and I've concluded that $D$ is decreasing for $x\in(0,4)$ and increasing for $x\in[4,\infty)$.
Since $D(4)>0$ for monotonicity I can conclude that the inequality is true in the interval $[4,\infty)$, but now I'm unsure on how study the thing in the interval $(0,4)$.
Since $D(4)>0$ it is enough to show that $D$ is positive in $(0,4)$, I would like to use the fact that $D$ is decreasing in $(0,4)$ but I can't calculate $D(0)$ since $D$ is not defined at $x=0$; so I would like to use the fact that $D$ tends to $\infty$ as $x \to 0^+$, by that it follows that for all $K>0$ there exists $\delta_K>0$ such that if $0<x<\delta_K$ it is $D(x)>K>0$.
So for $x\in(0,\delta_K)$ it is $D(x)>0$, then $D$ is decreasing and since $D(4)>0$ it follows that $D$ is positive in the interval $(0,4)$ as well. Is this correct?
Another question, an alternative approach was the following: since $x>0$ I thought about letting $x=t^2$, so the statement would be equivalen to $t\geq 2 \ln t$; the point is that I'm not sure if this is valid. I think it is valid because for $x\in(0,\infty)$ the map $x\mapsto t^2$ is bijective and so I don't lose informations, so solving the inequality in $t$ the other set where $t$ varies (which in this case is the same of the initial set) is equivalent of solving it in the initial set. Is this correct? I mean, if I do these kind of substitutions, I have to check that they are invertible? Thanks.

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