a problem of mu , pi , epsilon (journal)

Let $p_n$ denotes the $n$ th prime and $\pi(n)$denotes the numbers of primes $\leq$ $n$ . Prove that for $n$ $\geq$ $6$ the $inequality$ holds : $\pi(\sqrt{\displaystyle{\prod_{i=1}^{n}}}p_i)$ $>$ $2n$ ($mu$ $pi$ $epsilon$ $journal$)

Solution :Processing by induction for $n = 6$ ,we have $\pi(\sqrt{\displaystyle{\prod_{i=1}^{6}}}p_i)$ = $\pi(\sqrt 30030)\geq \pi(50)$ = 15 $>$ 6$\cdot$2.

So , this is valid for n = 6.
Now suppose $\pi(\sqrt{p_1p_2..........p_n)} > 2n$ for some n$\geq$ 6.
We will show that $\pi(\sqrt{p_1p_2..........p_np_{n+1})} > 2(n+1) = 2n+2 $. That is ,
$\pi(\sqrt{p_1p_2..........p_n}\cdot{\sqrt{p_{n+1}}}$) $ > 2n +2 $
Since n$\geq$ 6 , then $p_{n+1} \geq p_7 = 17$ and $\sqrt{p_{n+1}} > 4 $.
Now , it is easy to say that $\pi(\sqrt{p_1p_2..........p_n}\cdot{\sqrt{p_{n+1}}}$) $\geq$ $\pi(4\sqrt{p_1p_2..........p_n}$

Now, $Bertrand's Postulate$ guarantees a prime number between $k$ and $2k$ for all $k > 1.$ Thus $\pi((4k)-\pi(2k)$)and $\pi((2k)-\pi(k)$ )are each atleast $1$ , which follows that $\pi((4k)-\pi(2k)$)+ $\pi((2k)-\pi(k)$ )$\geq1+1$.
Now observe that ,$\pi(4k)$ = $\pi((4k)-\pi(2k)$)+ $\pi((2k)-\pi(k)$ )+$\pi(k)$.
Then
$\pi(\sqrt{p_1p_2..........p_np_{n+1})}$ $\geq$ $\pi(4\sqrt{p_1p_2..........p_n}$ =
$\pi(4\sqrt{p_1p_2..........p_n}$$\pi(2\sqrt{p_1p_2..........p_n}$ + $\pi(2\sqrt{p_1p_2..........p_n})$)- $\pi(\sqrt{p_1p_2..........p_n}$)$>1+1+2n$ ( since ,$\pi(\sqrt{p_1p_2..........p_n)} > 2n$)//That is $\pi(\sqrt{p_1p_2..........p_np_{n+1})}$ $>$ $2n+2$.
So, we are done .

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