Proving that $\pi(2x) < 2 \pi(x) $ 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.
 A: 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):

A: While this isn't my main area of study, I do think we are stuck with proving $Li(2x)<2Li(x)$ (ignoring $O(x^{\theta}\ln(x))$) for large $x$.
The definition of the offset logarithmic integral is
$$Li(x)=\int_2^x\frac{dt}{\ln(t)}$$
We want to take this and prove that $Li(2x)<2Li(x)$.
Since $\frac{d}{dx}Li(x)=\frac1{\ln(x)}>0$, we see that the offset logarithmic integral has a positive derivative.  Since $\frac{d}{dx}\frac1{\ln(x)}=\frac{-1}{x(\ln(x))^2}<0$, we know this function is concave.
This implies that for large enough $x$, $Li(2x)<2Li(x)$ since $Li(x)$ is concave.  I'm not quite sure how to do a formal proof or whatever, but I imagine we can go along those lines.
If $Li(2x)<2Li(x)$, I would think
$$\frac{d}{dx}Li(2x)<\frac{d}{dx}2Li(x)$$
$$\frac2{\ln(2x)}<\frac2{\ln(x)}$$
And the last line is much easier to prove.
