A particular cases of second Hardy-Littlewood conjecture

Question

Can someone solve this problem?

(or does someone known a proof of this problem, if it exists?)

for every $n\geq 2$, $\pi(2n)\leq 2\pi(n)$.

Thanks a lot!

Answer 1

By the prime number theorem, \[2\pi(x) - \pi(2x) = 2 \int_{2}^{x} \frac{dt}{\log t} - \int_{2}^{2x} \frac{dt}{\log t} + O\left(x\exp(-c\sqrt{\log x})\right).\] By making the change of variables $t \mapsto t/2$ in the second integral, the two integrals become \[2 \int_{2}^{x} \frac{dt}{\log t} - 2 \int_{1}^{x} \frac{dt}{\log 2t},\] which is \[2 \log 2 \int_{2}^{x} \frac{dt}{\log t(\log t + \log 2)} - 2 \int_{1}^{2} \frac{dt}{\log t + \log 2}. \] Using integration by parts, one can show that this is asymptotic to $2 \log 2 \frac{x}{(\log x)^2}$, and so \[2\pi(x) - \pi(2x) \sim 2 \log 2 \frac{x}{(\log x)^2}.\] So it is certainly true that $2\pi(x) \geq \pi(2x)$ for all sufficiently large $x$.