Werner D. Sand sent the following conjecture from his own
invention:

Let p, q, r be three consecutive prime numbers, p<q<r. Then
1/p < 1/q + 1/r.

Question: Prove it or show it false.

Contributions came from Farideh
Firoozbakht, J. K. Andersen, Jean Brette, Fred Schneider, Andrew Rupinski &
Joseph L. Pe. At the end this 'conjecture' resulted easy to solve,
nevertheless the proof came in distinct envelopes, as you will see.

Farideh, Andersen, Fred and Andrew sent
proofs based in the same concept: there at least two primes between p & 2p:

For p>5 we get pi(2p) - pi(p) >= 2,a
result by Ramanujan,
http://mathworld.wolfram.com/RamanujanPrime.html

This means there are at least 2 primes in [p+1,...,2p].

If p, q, r are consecutive primes then
both q and r must be in [p+1,...,2p].

q < 2p implies 1/(2p) < 1/q.

r < 2p implies 1/(2p) < 1/r.

1/p = 1/(2p) + 1/(2p) for any value of p.

Combining this gives the wanted 1/p = 1/(2p) + 1/(2p) < 1/q + 1/r.

Pe's attack came from a distinct side:

Conjecture #43 is true. Here is a
proof:

The conclusion follows from these two propositions:

(1) Let a(n) be a sequence of positive integers such that

a(n+1) < phi*a(n) for all n; here phi = (1+sqrt(5))/2 is the golden ratio.
Then 1/a(n) < 1/a(n+1) + 1/a(n+2).

(2) For any positive integer n > 48, there is a prime between n and
(9/8)*n. (See the article on Bertrand's Postulate in "Prime Numbers" by D.
Wells".)

Because of (2) above, the primes > 48 satisfy the condition of (1), and so
1/p < 1/q + 1/r for consecutive primes p < q < r larger than 48. But it is
easily checked that the conclusion holds also for primes < 48. This shows
that conjecture # 43 is true.

It only remains to prove proposition (1) above.

Proof of (1): From the hypothesis, we have

a(n+1) < phi*a(n)

a(n+2) < phi*a(n+1) < (phi^2)*a(n),

so that

1/(phi*a(n)) < 1/a(n+1)

1/((phi^2)*a(n)) < 1/a(n+2).

Adding corresponding sides of the previous two inequalities,

(1/a(n))*[1/phi + 1/(phi^2)] < 1/a(n+1) + 1/a(n+2).

Since phi^2 - phi - 1 = 0, we get 1/phi + 1/(phi^2) = 1, so that the
expression in square brackets in the previous inequality equals 1.

Therefore,

1/a(n) < 1/a(n+1) + 1/a(n+2),

as required.

Brette argued from this side:

In fact , q<3p/2
suffices, as follow:

The conjecture 43 is true.

« *The Prime pages* » give some results on gaps betweeen consecutive
primes.

http://primes.utm.edu/notes/gaps.html

Let g(*p*) be the number of composites between *p* and the next
prime :

if *p* and *q* are consecutive primes, we have *q* =
*p* + g(*p*) + 1.

In the § 4, **Bounds on g(***p*), one can find an upper bound
for g(*p*), after Nagura (1952) « g(*pn*) < (1/5) *pn* for
*n* > 9 » , where *pn* is the *n*th prime

So, for * p* > 23, and *q* the next prime, we have *q*
< *p* + (*p*/5) + 1

or *q* < *p *+ (*p*/2) = 3*p*/2, which is true for
all primes p>7

Taking the inverse, we have (1/*q*) > 2/3*p*

Since *q* and *r* are also consecutive, we have *r*
< 3*q*/2 < 9*p*/4 and (1/*r*) > 4/9*p. *So (1/*q*)
+ (1/*r*) > 6/9*p* + 4/9*p* = 10/9*p* > 1/*p*.

***