Is $\frac{\sqrt{x}}{\ln(x)-1}$ strictly increasing for $x \ge 21$

Question

I was thinking the prime counting inequalities from Pierre Dusart. In particular, I was interested in thinking about when $x \ge 5393$

$$\frac{x}{\ln(x)-1} < \pi(x)$$

I was using Excel with $\dfrac{\sqrt{x}}{\ln(x)-1}$ which seemed to go up and down up until $x=21$.

Does $\dfrac{\sqrt{x}}{\ln(x)-1}$ strictly increase when $x \ge 21$?

My thinking is yes.

Here's my thinking:

The derivative of $\dfrac{\sqrt{x}}{\ln(x)-1}$ is:

$$\frac{\ln(x) - 3}{2\sqrt{x}(\ln(x) - 1)^2}$$

This is positive for $\ln(x) > 3$ so that as long as $x > e^3 \approx 20.086$.

This suggests, if I am doing my calculations correctly, that for $x \ge 5393$:

$$\pi(x) > \sqrt{x}$$.

My only reason for doubting my thinking is that I have not seen this inequality before.

Here's my argument for $\dfrac{x}{\ln(x)-1} > \sqrt{x}$ for $x \ge 5393$

(1) $\dfrac{\sqrt{5393}}{\ln(5393) -1 } > 1$

(2) If strictly increasing, then for $x \ge 5393$,

$$\frac{\sqrt{x}}{\ln(x)-1} > 1$$

(3) So that:

$$\pi(x) > \frac{x}{\ln(x)-1} > \sqrt{x}$$

Answer 1

Your argument that it is increasing for $x \gt 21$ is correct. The final conclusion is correct, too, but it is not a very tight bound. $\ln(x)-1$ is tiny compared to $\sqrt x$ as $x$ gets large.

Answer 2

First, we define $f(x)=\frac{\sqrt(x)}{ln(x)-1}$. Then, it follows that $$f'(x)=\frac{0.5x^{-1/2}(ln(x)-1)-x^{-1/2}}{(ln(x)-1)^2}=\frac{x^{-1/2}(0.5ln(x)-1.5)}{(ln(x)-1)^2}$$

So we can see that the sign of the derivative entirely depends on the sign of $(0.5ln(x)-1.5)$. Nevertheless, for $x\geq 21$, $ln(x)>3$, so that $(0.5ln(x)-1.5)>0$, and therefore, for $x\geq 21$ we have $f'(x)>0$, and the function is strictly increasing.