Show $\pi (X) \leq 2 \pi (X/2)$

Question

Let $\pi (X)$ be the prime counting function. How can I prove that, for large enough $X$, $\pi (X) \leq 2 \pi (X/2)$? I have tried applying the Prime Number Theorem, but cannot see a way to get the result from it. Thanks

Answer 1

By using the Prime Number Theorem, it follows that $$\lim_{X\to +\infty} \frac{\pi(X)}{\pi(X/2)}=\lim_{X\to +\infty} \frac{X/\ln(X)}{(X/2)/\ln(X/2)}=2\lim_{X\to +\infty} \frac{\ln(X)-\ln(2)}{\ln(X)}=2$$ which implies that for any $\epsilon>0$, there is $M_{\epsilon}>0$ such that for $X>M_{\epsilon}$, $$\pi (X) < (2+\epsilon) \pi (X/2).$$

For your inequality, you need a more refined analysis. See Proving that $\pi(2x) < 2 \pi(x) $