Proving that $π(2x)-π(x)>\dfrac{x}{2\ln(x)}$ if $x≥e^6$

Question

This is from CryptoSchool by Joachim von zur Gathen, exercise 3.10 (Finding prime numbers).

Prove $π(2x)-π(x)>\dfrac{x}{2\ln(x)}$ if $x≥e^6$

I can not to prove the previous proposition, where $\pi(x)$ denote the number of primes $p$ with $p \le x$ and $π(x)≈\dfrac{x}{\ln(x)}$

I would appreciate any help.

Answer 1

The Non-asymptotic bounds on the prime-counting function section of Wikipedia's "Prime number theorem" article states Pierre Dusart proved in $2010$, in Estimates of Some Functions Over Primes without R.H., that

\begin{aligned}{\frac {x}{\log x-1}}&<\pi (x)&&{\text{for }}x\geq 5393,{\text{ and}}\\\pi (x)&<{\frac {x}{\log x-1.1}}&&{\text{for }}x\geq 60184\end{aligned}

Thus, for $x \ge 60184$, these $2$ inequalities can be combined to give

$$\frac {x}{\ln(x)-1} \lt \pi(x) \lt \frac {x}{\ln(x)-1.1} \tag{1}\label{eq1A}$$

Next, note you have

$$\ln(x) \gt 3.3 \implies \ln(x) - 1.1 \gt \left(\frac{2}{3}\right)\ln(x) \tag{2}\label{eq2A}$$

$$\ln(2) \lt 1 \implies \ln(2) - 1 \lt 0 \tag{3}\label{eq3A}$$

You have $\pi(2x) − \pi(x)$ is greater than the minimum for $2x$ in \eqref{eq1A} minus the maximum for $x$ in \eqref{eq1A}. This, along with \eqref{eq2A} and \eqref{eq3A}, gives

$$\begin{equation}\begin{aligned} \pi(2x) − \pi(x) & \gt \frac{2x}{\ln(2x) - 1} - \frac{x}{\ln(x) - 1.1} \\ & = \frac{2x}{\ln(x) + (\ln(2) - 1)} - \frac{x}{\ln(x) - 1.1} \\ & \gt \frac{2x}{\ln(x)} - \frac{x}{\left(\frac{2}{3}\right)\ln(x)} \\ & = \frac{4x}{2\ln(x)} - \frac{3x}{2\ln(x)} \\ & = \frac{x}{2\ln(x)} \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

Since $e^6 \approx 403.4$, you will need to do a check of the values $404 \le x \le 60183$, such as by using a computer program, to confirm \eqref{eq3A} applies for those values as well.

Answer 2

By the result of Rosser and Schoenfeld's Approximate formulas for some functions of prime numbers, we have that $\frac{x}{lnx} \leq \pi (x) \ \forall x \geq 17$. Does that help at all?