reciprocal consecutive primes
- 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.
H the harmonic mean
G the geometric mean
A the arithmetic mean
Then Cauchy says that H < G < A
Can you use that?
- --- theo2357 wrote:
>I think this is implied by the Iwaniec & Pintz finding, namely
> Let p,q,r be three consecutive primes. How can be proven that
> 1/p < 1/q + 1/r? It's NOT trivial!
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.