Help with a discrete math homework problem

[aporter1]

Homework Statement

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

Homework Equations

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.

 

[Mark44]

Homework Statement

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

Homework Equations

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

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.

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

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.

 

[aporter1]

okay so if i start with r<s, i have r<s.

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?

 

[dimension10]

Homework Statement

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

Homework Equations

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.

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

 

[Mark44]

okay so if i start with r<s, i have r<s.

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?

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?

 

[aporter1]

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

 

[Mark44]

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,

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

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

? You're not adding 2.

2r < 2(s+r) < 2s

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.

then,

r< (s+r)/2 < s

This doesn't follow from your previous inequality.

 

[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

 

[Mark44]

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.

 

[aporter1]

okay now where do i start with the second part?

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?

 

[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

 

[Mark44]

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)

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?

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

This doesn't make any sense to me.

 

[aporter1]

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

 

[Mark44]

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

 

[aporter1]

so how do i go about doing that then? I have to start with:

2/[(1/r)+(1/s)]²

 

[Mark44]

so how do i go about doing that then? I have to start with:

2/[(1/r)+(1/s)]²

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

As before, start with the given statement: r < s

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?

 

[aporter1]

its greater than zero because there are fractions

 

[aporter1]

would i add r²+s² to both sides?

 

[Mark44]

its greater than zero because there are fractions

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.

would i add r²+s² to both sides?

Yes.