Understanding the proof behind $\pi(x) \ge \frac{\log 2}{2}\frac{x}{\log x}$

Question

I am trying to understand the argument behind the proof that:

Given:

$$\pi(2n) \ge \log 2\frac{2n}{\log 2n}-1$$

Then for $x \ge 2$:

$$\pi(x) \ge \frac{\log 2}{2}\frac{x}{\log x}$$

Here's the argument:

1. The inequality can be checked directly for $x \le 16$
2. Pick an integer $n$ with $16 \le 2n < x \le 2n+2$
3. $\dfrac{2n}{\log 2n} - \dfrac{n+1}{\log 2n} = \dfrac{n-1}{\log 2n} \ge \dfrac{7}{4\log 2} \ge \dfrac{1}{\log 2}$
4. Hence: $$\pi(x) \ge \pi(2n) \ge \log 2\frac{2n}{\log 2n} - 1 \ge \frac{(n+1)\log 2}{\log(2n+2)} \ge \frac{\log 2}{2}\frac{x}{\log x}$$

I get lost at step #4. I don't understand why step #3 proves that:

$$\log2\frac{2n}{\log2n}-1 \ge \frac{(n+1)\log 2}{\log(2n+2)}$$

I also don't understand how we can be sure that:

$$ \frac{(n+1)\log 2}{\log(2n+2)} \ge \frac{\log 2}{2}\frac{x}{\log x}$$

If anyone can help me to understand these two steps, I would greatly appreciate it.

Thanks,

-Larry

Answer 1

In reverse order, the last step is from your step (2). Just multiply both sides by $2$ and compare numerator and denominator to $x$ as delimited in (2).

Step (4): Multiply (3) by $\ln 2. $ You get

$$\ln 2\frac{2n}{\ln 2n}-\frac{\ln 2(n+1)}{\ln 2n} \geq 1 $$

or

$$\ln 2 \frac{2n}{\ln 2n}-1 \geq \frac{\ln 2(n+1)}{\ln 2n}\geq \frac{\ln 2(n+1)}{\ln (2n+2).}$$