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Math Help - Inequality with four consecutive prime numbers

  1. #1
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    Inequality with four consecutive prime numbers

    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...
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  2. #2
    mfb
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    Re: Inequality with four consecutive prime numbers

    As 8 does not depend on n, it would be sufficient to show pn>8 for n>=3.
    This is wrong, but the few exceptions can be checked manually.
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  3. #3
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    Re: Inequality with four consecutive prime numbers

    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.
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  4. #4
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    Re: Inequality with four consecutive prime numbers

    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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