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

Question

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.

Answer 1

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 $n\gt 6$, $p_n\gt 16\Rightarrow \sqrt{p_n}\gt 2^2$.