Is it possible to prove that the number of prime numbers between $0$ and $n$ is greater than the number of prime numbers between $n$ and $2n$, for all $n > 10$? $$\pi(n)>\pi(2n)-\pi(n)$$
By the prime number theorem, we know that:
$$\lim_{n\rightarrow\infty}\frac{\ln(n)\pi(n)}{n}=1$$
so:
$$\lim_{n\rightarrow\infty}\pi(n)=\frac{n}{\ln(n)}$$ $$\lim_{n\rightarrow\infty}\pi(2n)=\frac{2n}{\ln(2n)}$$
$$\lim_{n\rightarrow\infty}\frac{\pi(n)}{\pi(2n)}=\frac{n\ln(2n)}{2n\ln(n)}=\frac{\ln(2n)}{2\ln(n)}$$
We know that:
$$\lim_{n\rightarrow\infty}\frac{ln(2n)}{2ln(n)}=\frac{1}{2}$$
therefore:
$$\lim_{n\rightarrow\infty}\frac{\pi(n)}{\pi(2n)}=\frac{1}{2}$$ $$\lim_{n\rightarrow\infty}2\pi(n)=\pi(2n)$$
$$\lim_{n\rightarrow\infty}\pi(n)=\pi(2n)-\pi(n)$$
This suggest that my inequality is false for $n=\infty$ because: $$\pi(\infty)=\pi(2\infty)-\pi(\infty)$$
Also, someone told me that since the relation between $\pi(n)$ and $\frac{n}{\ln(n)}$ is asymptotic, there are infinitely many n for which: $$\pi(n)<\pi(2n)-\pi(n)$$
but except for $n=2,4,10$, where the difference between $\pi(n)$ and $\pi(2n)-\pi(n)$ is $0$, the result of $\left(\pi(n)\right)-\left(\pi(2n)-\pi(n)\right)$ seems to grow slowly as n grows. Since the density of primes decrease as n grows, i don't understand how there could be more prime numbers between $n$ and $2n$ than between $0$ and $n$...
We need some nonasymptotic bounds. To get both of the below bounds, we need $x\ge 60184$; these were proved by Dusart in 2010.
$$\frac{x}{\log x -1}<\pi(x)<\frac{x}{\log x-1.1}$$
Hence, $$2\pi(x)-\pi(2x)>\frac{2x}{\log x-1}-\frac{2x}{\log(2x)-1.1}=2x\left(\frac{1}{\log x-1}-\frac{1}{\log x+\log 2 -1.1}\right)$$ Because $\log 2-1.1\approx -0.4$, $$\log x-1 < \log x +\log 2-1.1$$
Hence the RHS is positive, and the OP's claim is true for $x\ge 60184$; we leave $10\le x\le 60184$ for someone willing to do the computation.
Note: the result holds because $2>e^{0.1}\approx 1.105$. Hence, it can be strengthened, by replacing $2$ by a smaller constant, e.g. $$1.2\pi(x)-\pi(1.2x)>0$$ (holding for all $x\ge 60184$)