TITLE: Proving that $\pi(2x) < 2 \pi(x) $
QUESTION [7 upvotes]: In our analytic number theory class we were given the following problem as homework: prove rigorously that for large $x$ the number of primes in $(1,x]$ exceeds that in $(x,2x]$.
In class we proved the prime number theorem, and then proceeded to prove several results such as $\pi(x) = Li(x) +O(x^\theta \ln x)$ and the explicit formula for $\psi_1(x)$. 
This is clearly quite intuitive but I'm lost as to what I can use to prove the result. Any help is greatly appreciated.

REPLY [6 votes]: It can be shown that for $x \ge 11$, $\pi(2x) < 2\pi(x)$

For $x=1$, it is not true since $2\pi(1) = 0$ but $\pi(2)=1$
For $x \in \{2,4,10\}$, $2\pi(x)=\pi(2x)$

Pierre Dusart showed that for $x \ge 60184$
$$\frac{x}{\ln{x}-1} < \pi(x) < \frac{x}{\ln {x}-1.1}$$

For the proof, see Theorem 6.9 here.

For $a \ge 2$, $\ln ax - 1.1 = \ln a + \ln x - 1.1 \ge \ln 2 - 1.1 + \ln x > \ln x - 0.41 > \ln x - 1$
So, it follows that for $x \ge 60184, a \ge 2$:
$$\pi(ax) < \frac{ax}{\log{ax}-1.1} < \frac{ax}{\log x - 1} < a\pi(x)$$
By brute force, it can be shown that in all cases where $x < 60184$, $\pi(2x) < 2\pi(x)$.
Here is java code that I used to verify it (adding an image only):