Let's denote $\theta(x)$ as the sum of natural logarithms of primes up to $x$ and denote $\pi(x)$ as the number of primes up to number $x$.Does the inequality

$$\pi(2x)-\pi(x) \geq \frac{\theta(2x)-\theta(x)}{\log(x)}$$

holds where log is natural logarithm?

Edit: Grammar fixes.

  • $\begingroup$ Is it donate or denote $\endgroup$ – Darkrai Nov 29 '17 at 18:02
  • $\begingroup$ @Greninja thanks, I edited it. $\endgroup$ – Bright Chancellor Nov 30 '17 at 16:55

Let $a_n = 1_{n \text{ is prime}}$ then $$\pi(2x)-\pi(x) = \sum_{n \in (x,2x]} a_n \le \sum_{n \in (x,2x]} a_n \frac{\log n}{\log x}= \frac{\theta(2x)-\theta(x)}{\log x}$$


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