Inequality with four consecutive prime numbers

**Szuman** · June 26th 2012, 02:17 AM

Hi,

I want to show the following:

Let $n \geq 3$ , $p_n$ - n-th prime number. Then $p^2_{n+3} < p_n p_{n+1} p_{n+2}$ .

My line of reasoning:

$p_{n+3} < 2 p_{n+2} \Rightarrow p^2_{n+3} < 4 p^2_{n+2}$

$4 p^2_{n+2} < 4 p_{n+2} \cdot 2 p_{n+1}$

So:

$p^2_{n+3} < 8 p_{n+2} p_{n+1}$

What next? I guess, I have to use $p_n < 2^n$

I will be very grateful for any help...

**mfb** · June 26th 2012, 08:02 AM

As 8 does not depend on n, it would be sufficient to show p_n>8 for n>=3.
This is wrong, but the few exceptions can be checked manually.

**richard1234** · June 26th 2012, 08:54 AM

$p_{n+3} < 8p_{n+1}p_{n+2}$ . Here, if you can show that $8 < p_n$ , you are done. This is not true for n = 3 or n = 4, but just like mfb said, you can check those cases separately.

**Szuman** · June 26th 2012, 11:07 AM

Thank you both for your kind replies.
That's exactly like you say. Of course, it's perfectly ok to check these two exceptions manually, however, I am wondering how to avoid it. Can you see any way to finish it more 'formally'?

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