# Proving $\pi(\sqrt{p_{1}p_{2}\cdots p_{n}})>2n$ for $n \geq 6$

I am having trouble in solving the following problem.

• Let $p_{n}$ denote the $n$-th prime. Then prove that $$\pi(\sqrt{p_{1}p_{2}\cdots p_{n}})>2n$$ for $n \geq 6$.

No idea how to start.

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What is the source of the problem? You should probably define $\pi$ to be safe. – Jonas Meyer Feb 6 '11 at 17:53
$\pi$ when used as a function in number theory is always the prime counting function, i.e. $\pi(x)$ is the number of primes less than or equal to $x$. Asking to have it defined here really shouldn't be needed – kahen Feb 6 '11 at 20:36
@kahen: Fair enough, it is true that I did not need to ask. However, on a general site like this it has the potential to lead to confusion, and giving definitions may increase educational value to some readers. There was previously an answer posted interpreting the LHS as the number $\pi$ multiplied with the square root. – Jonas Meyer Feb 6 '11 at 20:49
@Jonas: Source math.muni.cz/~bulik/vyuka/pen-20070711.pdf – anonymous Feb 6 '11 at 23:40
@Chandru1: Thanks very much for that. Looks like an interesting collection. – Jonas Meyer Feb 7 '11 at 4:01

The case $n=6$ can be verified by hand (you only need to find $13$ primes less than $\sqrt{2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13}$), and once you have it this follows by induction and Bertrand's postulate, because for $n\gt 6$, $p_n\gt 16\Rightarrow \sqrt{p_n}\gt 2^2$.
For the $n = 6$ case, why do you need to find $13$ primes? Do you maybe mean that $p_6 = 13$? – Alex Basson Feb 6 '11 at 19:23
I mean you need to find $13$ primes less than $\sqrt{2\cdot 3\cdots 13}$, because $13\gt 2\cdot 6$. I didn't think about the fact that that might be confusing with $p_6=13$. – Jonas Meyer Feb 6 '11 at 19:28