Prove that $\log x<\sqrt{x}$ for $x\geq 1$

Question

Prove that $\log x<\sqrt{x}$ for $x\geq 1$

Let $f(x)=\sqrt{x}- \log x$. So, $f(1)=1>0$. $f'(x)=\frac{1}{2\sqrt{x}}-\frac{1}{x}>0$ only when $x>4$. When I draw the graph of $f$ in Wolframalpha, it shows the result, but how do I prove it rigorously? Can someone please help?

Answer 1

So $f'(x) = 0$ exactly when $x = 2\sqrt{x}$. That is $x^2 = 4x$, so $x = 0$ or $x = 4$. (Note $f'(x)$ is not defined at $x=0$). The solution $x=4$ is a minimum and here $f(4) = \sqrt{4} - \log(4)>0$ (this can easily be seen using the second derivative test). So if the minimum is greater than $0$, then $f$ is always greater than $0$.