# Help with a discrete math homework problem

1. Jun 4, 2012

### aporter1

1. The problem statement, all variables and given/known data
show that for positive r and s, with r<s, we always have:

r<(r+s)/2<s and 2/[(1/r)+(1/s)]^2< 2rs< (r+s)^2

2. Relevant equations

3. The attempt at a solution
i have shown that r<s because r+r<s+r, 2r<s+r, r<(s+r)/2 and 2r< (2s+2r)/4
(r+s)^2= r^2+2rs+s^2 where are<r^2+2rs+s^2

not sure if I'm even on the right track or if i have even finished the problem. I'm kinda confused.

2. Jun 4, 2012

### Staff: Mentor

You've wandered off the right track. You don't need to (and shouldn't) show that r < s. You are given that r < s.

Some of what you wrote makes sense, but a lot doesn't (e.g., "(r+s)^2= r^2+2rs+s^2 where are<r^2+2rs+s^2"), and it's not well organized.

For each part, you should start with r < s. For the first part, the last line should be r < (r + s)/2 < s. For the second part, the last line should be 2/[(1/r)+(1/s)]^2< 2rs< (r+s)^2. Each line between the start and end should make sense logically and mathematically.

3. Jun 5, 2012

### aporter1

then, how do i show r<s for r+s/2 ? Would I start with r+r<r+s, if so, how do i incorporate the 2?

Last edited: Jun 5, 2012
4. Jun 5, 2012

### dimension10

Just break up the inequality into 2 parts, that is it.

5. Jun 5, 2012

### Staff: Mentor

You don't show r < s for anything. You are to assume this. You need to show that r < (r + s)/2 < s.

What you started with was OK, but you need to take it a bit further.

r < s
=> r + r < s + r and s + r < s + s -- Do you understand why this follows from r < s?
=> ? What can you do with the two inequalities above?

6. Jun 5, 2012

### aporter1

Re: help with a discrete math homework problem

okay so,

r + r < s + r < s + s, this is true because if r < s then, r < s +r and because r < s then r+r<s, so we have what follows,

r + r < s + r < s + s , now i believe if we add 2 ,

2r < 2(s+r) < 2s then,

r< (s+r)/2 < s

7. Jun 5, 2012

### Staff: Mentor

Re: help with a discrete math homework problem

Here's a better way to say it.
r < s (given)
r + r < s + r (by adding r to both sides of the given inequality)

r + s < s + s (by adding s to both sides of the given inequality)

Combining these, we get
r + r < r +s < s + s
No. The expressions at either end are right, since r + r = 2r and s + s = 2s, but you can't just stick a factor of 2 in the middle.

8. Jun 5, 2012

### aporter1

how to i prove the middle part then? I understand the other part now.

would i just divide out by two to get,

r<(r+s)/2<s

9. Jun 5, 2012

### Staff: Mentor

Yes.
2r < r + s < 2s ==> r < (r + s)/2 < s

What this is saying is that if r < s, the average of r and s will be between r and s.

10. Jun 5, 2012

### aporter1

2/[(1/r)+(1/s)]^2< 2rs< (r+s)^2

Would I start by supposing r<s again? and then,

would i start by expanding out (r+s)^2?

Last edited: Jun 5, 2012
11. Jun 5, 2012

### aporter1

so i've tried to figure it out and this is what i have come up with,

2rs< r2+2rs+s2, (by expansion of (r+s)2)

2/(2/rs) < 2rs, (by expansion of (1/r + 1/s)2) and finding a common denominator

12. Jun 5, 2012

### Staff: Mentor

Starting from r < s, how do you get to this?

When you expand (r + s)2, you get r2 +2rs + s2. IOW, (r + s)2 = r2 +2rs + s2.

How did you get an inequality out of that?

This doesn't make any sense to me.

13. Jun 5, 2012

### aporter1

well I was just comparing it to the other part in the problem

14. Jun 5, 2012

### Staff: Mentor

You have to be able to justify each step. Each inequality you write has to be a direct consequence of the one before it.

15. Jun 5, 2012

### aporter1

2/[(1/r)+(1/s)]2

16. Jun 5, 2012

### Staff: Mentor

The second part is quite a bit easier; namely, showing that 2rs < (r + s)2

Notice that r2 + s2 > 0 -- can you explain why?
What can you add to the left side (and the right side) to make the left side (r + s)2?

17. Jun 5, 2012

### aporter1

its greater than zero because there are fractions

18. Jun 5, 2012

### aporter1

would i add r2+s2 to both sides?

19. Jun 6, 2012

### Staff: Mentor

No, that's not why. If r < s, it does not necessarily follow that 1/r + 1/s > 0. Consider r = -1 and s = 2. We have r < s, but 1/r + 1/s = 1/(-1) + 1/2 = -1/2 < 0.
Yes.