# 1/p + 1/q + 1/r

1. Jan 10, 2005

### recon

What is the maximum sum less than 1 but more than 0 that can be formed from $$\frac{1}{p} + \frac{1}{q} + \frac{1}{r}$$, where p, q and r are positive integers?

I posted this problem in the homework forum even though it isn't homework. Since I have not yet received any help, I have posted it here.

2. Jan 10, 2005

### Zurtex

I'm not sure there is a solution.

3. Jan 10, 2005

### recon

Is there a way of proving this then?

4. Jan 10, 2005

### Hurkyl

Staff Emeritus
Brute force doesn't work?

5. Jan 10, 2005

### jcsd

1/2 +1/4 + 1/5 = 19/20

edited to add: yes brute force is the way, but you onluy need actually consider about 5 or so different sums once you've excluded the ones thta cannot possibly be the correct answer. For example the fact that 1/2 + 1/4 + 1/5 > 1/4 + 1/4 + 1/4 implies that at least one term must be greater than 1/4.

Last edited: Jan 10, 2005
6. Jan 10, 2005

### Zurtex

If you rearrange this you can show you want pq + qr + rp to be as close as possible to pqr but such that pq + qr + rp < pqr. Can't say my maths is good enough to think about 3 independent variables like this.

But it would seem to me if you could up with some functions p(t), q(t) and r(t) such that p(t)q(t) + q(t)r(t) + r(t)p(t) monotonically increases and so does p(t)q(t)r(t) and that p(t)q(t) + q(t)r(t) + r(t)p(t) < p(t)q(t)r(t) holds true for all $p(t), q(t), r(t) \in \mathBB{N}$ then that is proof there is no solution.

Not very useful I know I'm sorry.

7. Jan 10, 2005

### Hurkyl

Staff Emeritus
I can do better!

8. Jan 10, 2005

### Zurtex

p=7
q=3
r=2

Is closer

9. Jan 10, 2005

### Gokul43201

Staff Emeritus
How about $$\frac{1}{2} + \frac{1}{3} + \frac{1}{7}~~?$$

Edit : didn't see zurtex's post.

Last edited: Jan 10, 2005
10. Jan 10, 2005

### jcsd

hmm lets try again:
oops yes, I actually did consider: 1/2 + 1/3 + 1/7 = 41/42, but I added it up wrong

Here's the proof then tthat this is indeedf the largets possible answer.

As I said earlier we already know that at least one term must be greater than 1/4:

2/3 + 2/3 + 1/4 < 1/2 + 1/3 + 1/7 therefore at leats one term muct be a 1/2

As 1/2 + 1/3 + 1/6 > 1 and 1/2 + 1/3 + 1/7 < 1, the rest of the terms must be less than or equal to 1/3 and gretare than or equal to 1/7

which means only the sums:

1/2 + 1/3 + 1/7

1/2 + 1/4 + 1/5

1/2 + 1/6 + 1/6

need be considered of them the greaest is 1/2 + 1/3 + 1/7.

edited again to add: looks like Gokul beat me to it.

Last edited: Jan 10, 2005
11. Jan 10, 2005

### Gokul43201

Staff Emeritus
Re-write the problem as follows :
For a given value of r, find p, q that maximizes the sum of reciprocals subject to it being less than 1.

If r> 7, need maximal 1/p + 1/q < (r-1)/r.

This is achieved with p=2, q=3 for all such r, since :
1/2 + 1/3 < 7/8 => 1/2 + 1/3 < (r-1)/r , for all r>7
AND
the next larger value of 1/p + 1/q = 1/2 + 1/2, which is clearly not allowed, as the sum will exceed 1.

Thus the maximal solution for any r>7 is 1/2, 1/3, 1/r

Clearly, any of these solutions is inferior to 1/2, 1/3, 1/7.

This leaves you with only having to check r=5,6.

Last edited: Jan 10, 2005
12. Jan 10, 2005

### recon

What should this equation really read like?

13. Jan 11, 2005

### Zurtex

I think:

1/3 + 1/3 + 1/4 < 1/2 + 1/3 + 1/7