Show there exists a $K > 0$ such that for all $x> K$ the interval $]x,2x]$ contains fewer primes than the interval $[0,x]$.
My approach so far has been to try to show $\pi(2x)-2\pi(x)$ is negative for all $x>K$ as this would be the desired conclusion. My idea is to apply the Prime Number Theorem, but so far I have not been able to do it in a fruitful way.
I would appreciate any help or hints for this problem. Thanks!
Use the prime number theorem in the form $$\pi(y) = \int_2^y \frac{1}{\log t}dt + O_A(y (\log y)^{-A}).$$ We have $$2\pi(x) - \pi(2x) = \int_2^x \frac{1}{\log t}dt - \int_x^{2x}\frac{1}{\log t}dt + O_A(x (\log x)^{-A}),$$ and we easily see that $$\int_x^{2x}\frac{1}{\log t}dt \leq \frac{x}{\log x}.$$ If we now integrate by parts once, we get $$\int_2^x \frac{1}{\log t}dt = O(1) + \frac{x}{\log x} + \int_2^x \frac{1}{\log^2 t}dt,$$ so $$2\pi(x) - \pi(2x) \geq (1+o(1))\int_2^x \frac{1}{\log^2 t}dt \gg \frac{x}{\log^2 x}$$ as $x$ tends to infinity.