- Actually, it doesn't seem that hard:

1/p < 1/q + 1/r

q>=p+2

r>=q+2>=p+4

It must be shown that 1/p < 1/(p+2) + 1/(p+4)

equivalently (p+2)(p+4)/p < 2p +6 or

p+6+8/p < 2p+6 or

8/p < p or

p^2 > 8 which happens when p>= 3

Notice it also works for 2, 3, and 5 since

1/2 < 1/3 + 1/5 ( 7.5 < 8 )

Milton L. Brown

> [Original Message]

> From: theo2357 <Theo.3.1415@...>

> To: <primenumbers@yahoogroups.com>

> Date: 11/2/2005 9:46:17 AM

> Subject: [PrimeNumbers] reciprocal consecutive primes

>

> Let p,q,r be three consecutive primes. How can be proven that

> 1/p < 1/q + 1/r? It's NOT trivial!

>

>

>

>

>

>

> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com

> The Prime Pages : http://www.primepages.org/

>

>

> Yahoo! Groups Links

>

>

>

>

> - On Wed, 2 Nov 2005, theo2357 wrote:

> Let p,q,r be three consecutive primes. How can be proven that

This is equivalent to a statment that for a given prime q the preceding

> 1/p < 1/q + 1/r? It's NOT trivial!

>

>

prime p>=2q/3.

Supose it is true. Moreover, it is evident (and there is threom regarding

it, I think) r<2q. These inequalities lead:

2 3

a) p>=-q => 1/p <=--

3 2q

1 2 1 3

b) r<2q => 1/r > -- => 1/q+1/r > -- + -- = -- >= 1/p

2q 2q 2q 2q

If p>=2q/3 is true, then r <= 3q/2. So 1/q+1/r >= 5/3q = 1.666...*(1/q)

whereas 1/p <= 1.5*(1/q).

So, my formulation is:

Show that for any prime p the next prime q is not larger than 1.5*p

(equality is valid for p=2, only).

Maybe it has been proven?

Wojtek

===============================================

Wojciech Florek (WsF)

Adam Mickiewicz University, Faculty of Physics

ul. Umultowska 85, 61-614 Poznan, Poland

Phone: (++48-61) 8295033 fax: (++48-61) 8257758

email: Wojciech.Florek@... - 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? - --- 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.