Prove: for $x$ large enough, there are more primes in the interval $(1,x]$ than in $(x,2x]$ I want to show that there are more primes in the interval $(1,x]$ than in $(x,2x]$, when $x$ is large enough. Let $\pi$ be the prime counting function, ie. $\pi(x)=\# \{ p \leq x | p \text{ is a prime} \}$. Then the above statement is equivalent to showing:
$\pi (2x)-2\pi(x)=-\log(2)\frac{x}{\log ^2x}+O\left(\frac{x}{\log ^3 x}\right)$
I have a hint saying to use the prime number theorem. I'm trying to use that $\pi(x)= \frac{x}{\log x} + \frac{x}{\log ^2 x} + O \left( \frac{x}{\log ^3 x} \right)$, but can't seem to rewrite it fitting. Maybe I need to use other aspects of the prime number theorem.
Added: It's useful to see alternative ways of solving the problem, but I'm still specifically seeking to show the equation above using PNT.
 A: From Dusart we know that $\frac{x}{\ln x}\left(1+\frac{1}{\ln x} \right) < \pi(x )<\frac{x}{\ln x}\left(1+\frac{1.3}{\ln x} \right) $ for all $x \geq 600$ and so you need to prove that $\pi(2x) -2 \pi(x) \leq \frac{2x}{\ln 2x} \left(1+\frac{1.3}{\ln 2x} \right)-2\frac{x}{\ln x}\left(1+\frac{1}{\ln x} \right) << 0  $
Or Equivalently $\frac{1}{\ln 2x} \left(1+\frac{1.3}{\ln 2x} \right) << \frac{1}{\ln x}\left(1+\frac{1}{\ln x} \right) => \ln x (1+\frac{1.3}{\ln 2x}) < (\ln x +\ln 2)(1+\frac{1}{\ln x}) => \ln x +\frac{1.3 \ln x}{\ln 2x} < ln x +1.3< \ln x+ 1 + \ln 2 < \ln x+1+\ln 2 + \frac{\ln 2}{\ln x} $
Checking for smaller cases the inequality is true for all $x\geq 11$
A: From the proof of the prime number theorem by the de la Vallée Poussin we have
$$\frac{x}{\log -(1-\epsilon)} < \pi(x) < \frac{x}{\log x -(1+\epsilon)}$$
for $x\ge x_{\epsilon}$. In other words, if we write $\pi(x) = \frac{x}{\log x - B(x)}$, then $B(x) \to 1$ ( see Legendre's constant).  In fact, we do have an explicit estimate of Dusart, but the above is enough.
We get
$\pi(2 x) < \frac{2x}{\log(2x) -(1+\epsilon)}$
and $\frac{2x}{\log x- (1-\epsilon) } < 2\pi(x)$
so we should have
$$\log (2x) -(1+\epsilon) > \log x - (1-\epsilon)$$
or $\log 2 > 2 \epsilon$. So we should take $\epsilon < \frac{\log 2}{2}$, and then get the inequality for $x \ge x_{\epsilon}$.
Note that in general for every $M>1$, we can get $\pi(Mx) < M \pi(x)$ for $x> \bar x_{M}$.
$\bf{Added:}$  Let us use the explicit Dusart bounds
$$\frac{x}{\log x -1} < \pi(x)$$
for $x\ge 5393$, and
$$\pi(x) < \frac{x}{\log x -1.1}$$
for $x \ge 60184$.  Therefore, for $x\ge 30 092$ we get
$$\pi(2x)< \frac{2x}{\log (2x) - 1.1} < \frac{2x}{\log x -1}< 2\pi(x)$$
Consider for  integer $4\le x \le 60183$,  the values of $\frac{\pi(x)}{ 2\pi(x/2)}$. It reaches $1$ at $4$, $7$, $8$, $9$, $13$, $19$, $20$, $21$, and values $< 1$ otherwise.
A: There is no need for error estimates.  Use the PNT in the form $\pi(x) \sim x/\log x$. That implies $\pi(2x)/\pi(x) \sim (2x/\log(2x))/(x/\log x) = 2(\log x)/\log(2x)$.  Now convince yourself that this tends to $2$ as $x\to \infty$, so the difference
$$
\pi(2x) - \pi(x) = \pi(x)\left(\frac{\pi(2x)}{\pi(x)} - 1\right)
$$
is large for large $x$.  Similar reasoning shows for each $c > 1$ that $\pi(cx) - \pi(x)$ is large for large $x$.
