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Let $\pi(x)$ be a prime-counting function (the function counting the number of prime numbers less than or equal to some real number $x$. For example $\pi(5)=3$, $\pi(4)=2$). Prove by elementary (high school-level) methods that there is a function $f: \mathbb{N} \to \mathbb{R}$ and there is $N_0 \in \mathbb{N}$ such that for all $n \ge N_0$ the inequality $\pi(n) \ge f(n)$ holds and $\lim \limits_{n \to \infty} \frac{f(n)}{\sqrt{n}}=\infty$.

My work. Well known that $\pi(n) > \frac{n}{\ln n}$. But the proof of this inequality is not elementary (high school-level). I searched, but did not find other estimate. For example, it would be a good idea to prove that $\pi(n) \ge n^{0.6}$ for all $n \ge N_0$.

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  • $\begingroup$ I ask about the argument which would be accessible to a high school student (olympiad problem). $\endgroup$ – Witold Jul 5 at 21:18
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    $\begingroup$ @XanderHenderson I think for the OP "elementary" means "not advanced" and "school" means "not university". The former does not modify the latter. $\endgroup$ – Ethan Bolker Jul 5 at 21:24
  • $\begingroup$ OK :) Not a university, but a very advanced school. $\endgroup$ – Witold Jul 5 at 21:27
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    $\begingroup$ The Chebyshev bound mentioned here is both elementary and strong enough for your requirements. $\endgroup$ – Momo Jul 5 at 21:47
  • $\begingroup$ @Momo thank you very much! $\endgroup$ – Witold Jul 6 at 12:30
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Here's a funny bound: $\forall m \geq 4 ,\pi(m) \geq \lfloor \log_2(\log_2(m)) \rfloor+1$. It's not a very good bound at all, but it does follow a very easy argument, akin to the usual proof of that there are infinitely many primes.

First, we use the easy-to-show fact that there must be another prime $p_2 < p_1^2 = 4$

Then, we generalize this, saying that there must be another prime $$p_{n+1} < \prod_{k=1}^np_k$$ And from our base cases $p_1=2$, $p_2<2^2$, we can apply this formula to get for that for $n \geq 1$, we have $$p_{n} < 2^{(2^{n-1})}\implies \log_2\log_2(p_n)+1<n$$ And finally this tells us that $\forall m\geq 4(>p_2)$, we have our funny bound!

If one proves/accepts Bertrand's postulate (that $\forall n>3, \exists p$ s.t. $p$ is prime and $n < p < 2n$), then one can use a similar argument to get a slightly less silly bound around $\log_2(n)$.

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