reciprocal consecutive primes

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• 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 9:07 AM
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?
• ... 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 2:20 PM
--- 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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