I have a question about a proof I read, linked here, a proof that $\sqrt{x}>\ln x$.
The second answerer considers the derivative of the function,$f\left(x\right)=\sqrt{x}-\ln x$. Now the derivative is negative for $0\leq x<4$. Does this not mean, that $\sqrt{x}<\ln x$ on that interval? I.e. The derivative being negative suggests that the function is decreasing? Why is the global minimum analysed? I must be missing something important!
$\endgroup$ 02 Answers
$\begingroup$The derivative of $f(x)=\sqrt{x}-\ln(x)$ is $0$ at $x=4$. Also its second derivative at $x=4$ is positive, so this point must be the function's minima.$f(x)\ge f(4)\forall x\in (0,\infty)$. And $f(4)\gt 0$. So $f(x)\gt 0 \forall x\in(0,\infty)$.
$\endgroup$ $\begingroup$No. It only means that $\sqrt x$ grows slower that $\ln x$ in $[0,4)$, but maybe $\sqrt x$ has started growing from a point above $\ln x$ (which is the case here; $\sqrt x$ starts from $0$ and $\ln x$ starts from $-\infty$).
Here is a sketch
