That's asymptotically true cause p(n) ~ n / log(n) -->
--> p(n+1) / p(n) ~ (n+1)/n * log(n) / log(n+1) = 1 and therefore you can find a natural
n0 so that if n > n0, then |p(n+1)/p(n) - 1| < 1/2 --> p(n+1)/p(n) < 3/2 for every n > n0.

That is, for n "bigger enough", 1/r + 1/q > 1/p always holds.

But it seems to hold for every (p,q,r), not only for the "bigger" ones. q/p < 3/2 is a
simpler necessary condition that seems to hold if p>7. I don't know now how to prove it
without using analysis (to prove the trueness for every 3-uple of consecutive primes).
Maybe there's a simple and clever argument, I recommend to search for better boundings to
the distribution of primes than Bertrand's Postulate (I'm sorry, I'm sooo lazy...).

Nice observation, btw.

Regards. Jose Brox.

Jose Ramón Brox

... From: Milton Brown Actually, it doesn t seem that hard: 1/p =p+2 r =q+2 =p+4 It must be shown that 1/p

Message 2 of 3
, Nov 3 12:56 AM

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----- Original Message -----
From: "Milton Brown" <miltbrown@...>

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)

----------------------------------------

FALSE. If r>=p+4, then 1/r <= 1/(p+4). Considering the case where 1/r = 1/(p+4) and 1/q =
1/(p+2) is a best case study, not a worst case one.

Jose Brox

Jose Ramón Brox

... From: Jan van Oort p^2 mn which is true iff ( m a ) ( condition i ) ...

Message 3 of 3
, Nov 3 4:00 AM

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----- Original Message -----
From: "Jan van Oort" <glorifier@...>

p^2 > mn

which is true iff

( m < p AND n < p ) OR ( m = p + a AND n = p - b AND b > a ) ( condition i )