Consider the theta function, $\theta(x)$=$\sum_{p\leq x} \log p$.
Lemma: $\theta(x)$ is strictly monotone increasing.
We thus have that $\theta(x)$<$\theta(2x)$. Note also that $\theta(2x)$=$\theta(x)$+$\sum_{p<q\leq2x} \log p$. So there is a q in between x and 2x. Assume there did not exist such a q. Then by the monotone condition, one has $\theta(x)$<$\theta(2x)$=$\theta(x)$+$\sum_{p<q\leq2x} \log p$. But by assumption, $\sum_{p<q\leq2x} \log p$=0. Thus we have that $\theta(x)$<$\theta(x)$. A contradiction. Thus we have a prime q, such that x<q$\leq$2x.
My only dilemma is the proof of the monotone condition. I can't seem to find it. But once I do, then this would imply Bertrand's postulate. I am sorry if I am not very good with these types of things. I just tried my best to find the most elementary proof of this conjecture.
So I would like to know if; 1) my proof is correct, 2) help with the proof of the lemma.
EDIT: The above lemma is false. Let me restate it:
Lemma: $\theta(x)$<$\theta(2x)$. Now I need help with this lemma. :/
The correct formula is $$\theta(2x) = \theta(x) + \sum_{x<p\le 2x}\log p$$ Then $\theta(2n) > \theta(n)$ is equivalent to say that there is a prime in $(n,2n]$. However, $\theta(x)$ is NOT strictly increasing: $\theta(7)=\theta(8)=\theta(9)=\theta(10)$.