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The question is as follows:

How can I prove the following theorem? $$\forall n \in \mathbb{N}, \ \pi(n) \geqslant \frac{\log n}{2\log 2} \tag1 \label1$$

I have no idea where to begin. I came across this equation on a paper that apparently proved Legendre's Conjecture. You can review the paper here, which introduces the open conjecture, but it only reveals two proof even though the title states there are three proofs. In the paper, it mentioned that a demonstrated proof of $\eqref1$ was on pp. $21$ but for some reason, the paper that I observed, and that of which is accessible in the link, holds only $7$ pages.

May somebody please provide a proof of $\eqref1$? I am just curious.

Thank you in advance.

Edit:

When I was about to post this question, I was unaware that such a similar question has already been posted and answered. Sorry about that.

Mr Pie
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    See here or here for example. Note that it is a ridiculously weak lower bound. – Daniel Fischer Jan 13 '18 at 12:57
  • @DanielFischer Thank you very much. You have given me more insight into this theorem. The second link I found most helpful, as the proof was very clear and I could at least participate in completing the proof by deriving this theorem from the inequality $n\leqslant 2^{\pi(n)}\sqrt{n}$ – Mr Pie Jan 13 '18 at 13:11
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    And, just to mention it, if a paper uses $(1)$ (in any other way than to showcase a slick proof), it is most likely not proving Legendre's conjecture. – Daniel Fischer Jan 13 '18 at 13:32
  • @DanielFischer So then how can we explain the title of the paper? – Mr Pie Jan 13 '18 at 13:34
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    I haven't looked at the paper, and the link goes to .../S63.pdf which doesn't say much. If the title of the paper claims it proves Legendre's conjecture, it's most likely that the author made an error. It is of course also possible that my heuristic was wrong and the paper indeed proves Legendre's conjecture. I wouldn't bet on that, though. – Daniel Fischer Jan 13 '18 at 13:42
  • Okay, I took a look, and in the first "proof" (section 3.1), the author uses the inequality $$a - b + \sum_{k = u}^v c_k - \sum_{k = w}^x c_k \geqslant a - b + \sum_{k = u}^v \min { c_m : m \geqslant 5} - \sum_{k = w}^x \min { c_m : m \geqslant 5}$$ which is of course invalid. In the subtracted sum, replacing $c_k$ with the maximum instead of the minimum would produce a valid - but entirely unhelpful - inequality. We have $\min { -c_m : m \geqslant 5} = - \max { c_m : m \geqslant 5} \neq -\min { c_m : m \geqslant 5}$. – Daniel Fischer Jan 13 '18 at 14:27
  • The second "proof" (section 3.2) makes the same mistake, and the third (section 4) too. That says just $$\pi(x) - \pi(y) \geqslant \frac{\log x}{2\log 2} - \frac{\log y}{2\log 2}$$ for $x = (n+1)^2$ and $y = n^2$. Of course that argument would show that $\pi(x) > \pi(y)$ for all $x > y$ if it were valid. – Daniel Fischer Jan 13 '18 at 14:35
  • In a nutshell: The author of that paper deduces $A - B \geqslant a - b$ from $A \geqslant a$ and $B \geqslant b$. – Daniel Fischer Jan 13 '18 at 19:37

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