# Homework Help: Write with at least one less abs value symbol

1. Feb 3, 2012

1. The problem statement, all variables and given/known data
Here we go. Write the following with at least one less absolute value symbol:

(ii) |(|a+b| - |a| - |b|)|

(iii) |(|a + b| + |c| - |a + b + c|)|

(iv) |x2 - 2xy + y2|

2. Relevant equations

12 Properties of Numbers.

3. The attempt at a solution

Let's just look at (ii) for now since I am getting all flustered with it:

|(|a+b| - |a| - |b|)|

There are 2 potential abs value symbols that could be dealt with:

1) I could try to show that the quantity |a+b| - |a| - |b| ≥ 0 and the outermost symbols could be dropped.

OR

2) Try to show that -|a|-|b| = -|a+b| and re-write the original expression as |(|a+b| - |a + b|)|

Either way is a pain. It seems like there are so many 'cases' to test. Am I correct in saying that I would need to test all of the following cases?

I. a = b > 0
II. 0 < b < a
III. 0 < a < b
IV. a < 0 < b where |a| < |b|
V. b < 0 < a where |b| < |a|
VI. a < 0 < b where |b| < |a|
VII. b < 0 < a where |a| < |b|
VIII. a = b = 0
IX. a = b < 0

Am I overkilling here? Are any of these equivalent? I am thinking "no" because its subtraction involved ....

2. Feb 3, 2012

### SammyS

Staff Emeritus
For (ii) What does the triangle inequality tell you about how |a+b| and |a|+|b| are related?

For (iv) Factor x2 - 2xy + y2 .

3. Feb 4, 2012

From the triangle inequality: |a+b| ≥ |a| + |b| → |a+b| -|a| - |b| ≥ 0 ... jeesh.....nice catch SammyS!

EDIT: oof! that's backwards! Now I'm screwed. I need sleep. Back in morning!

For (iv)

I see that it is |(x+y)(x-y)| but I am not sure what to do with that? Again there are many cases to test, right?

I was originally thinking that since x2 and y2 are always > 0, then I would want to try to show that: x2 + y2 ≥ 2xy for all x,y.

Last edited: Feb 4, 2012
4. Feb 4, 2012

### SammyS

Staff Emeritus
Not quite right!
(x+y)(x-y) = x2 - y2

Try again.

For (iii): Try the triangle inequality here too.

5. Feb 19, 2012

So I have come back to this one finally: I am still unsure if part (ii) ... I think I am missing the obvious.

(ii) $|(|a+b| - |a| - |b|)|$

From the triangle inequality I have that $|a+b|\le|a|+|b|\implies |a+b| - |a| - |b|\le 0$ so for sure I cannot remove the outer most abs value signs. This means that I am looking at removing (at least) one of the inner sets. As a guess, I am choosing to remove the signs around $|a+b|$ which means that I need to show that for any values a,b, that $|a+b| - |a| - |b| = a+b - |a| - |b|.$

But I don't know if this is correct. I am pretty sure it is not.

6. Feb 19, 2012

### LCKurtz

Are you really sure? What is $|x|$ if $x\le 0$?

7. Feb 19, 2012

I'm sorry LCKurtz, I don't think I understand what you're asking? |x|=-x if x≤0. But what is your point ?

8. Feb 20, 2012

### LCKurtz

Think about what you said above which I colored red. What if that is x?

9. Feb 20, 2012

Hi LKCurtz! I don't think I had enough foresight to realize that I could remove an abs value sign and rearrange the expression . How about this?

$|(|a+b|-|a|-|b|)|=|b|+|a|-|a+b|$

10. Feb 20, 2012

### LCKurtz

Yes. That's the idea.

11. Feb 21, 2012

Just to finish up:

(iii) $|(|a + b| + |c| - |a + b + c|)| = |a + b| + |c| - |a + b + c|$

since $|a + b + c| \le |a + b| + |c|$ by triangle inequality.

and
(iv) $|x^2 - 2xy +y^2| =x^2 - 2xy +y^2.$

Since, $x^2 - 2xy +y^2 = (x-y)^2 \ge 0 \text{ for all x.}$

12. Feb 21, 2012

### SammyS

Staff Emeritus
Yes. Those look good !