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reciprocal consecutive primes

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  • theo2357
    I don t know if it s proven. It s my own observation. 1/p qr/(q+r) which means p is greater than the half of the harmonic
    Message 1 of 5 , Nov 3, 2005
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      I don't know if it's proven. It's my own observation.
      1/p < 1/q + 1/r is equivalent to:
      p > qr/(q+r) which means p is greater than the half of the harmonic
      mean of q and r.
      Let
      H the harmonic mean
      G the geometric mean
      A the arithmetic mean

      Then Cauchy says that H < G < A
      Can you use that?
    • jbrennen
      ... I think this is implied by the Iwaniec & Pintz finding, namely that there is always a prime between x and x-x^(23/42), for any real number x 11. The
      Message 2 of 5 , Nov 3, 2005
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        --- theo2357 wrote:
        >
        > Let p,q,r be three consecutive primes. How can be proven that
        > 1/p < 1/q + 1/r? It's NOT trivial!
        >

        I think this is implied by the Iwaniec & Pintz finding, namely
        that there is always a prime between x and x-x^(23/42), for
        any real number x > 11.

        The Iwaniec & Pintz theorem can be used to show the weaker result,
        that for prime p >= 5, nextprime(p)/p < Phi, where Phi is the
        golden ratio, (sqrt(5)+1)/2 = 1.618...

        And if q/p < Phi, and r/q < Phi, then the original inequality
        1/p < 1/q + 1/r follows quite easily.
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