# Proving this inequality

1. Oct 23, 2013

### jimmyly

1. The problem statement, all variables and given/known data
|x + y| ≥ |x| - |y| [Hint: write out x = x + y - y, and apply Theorem 3, together with the fact that |-y| = |y|]

2. Relevant equations
Theorem 3: |a + b| ≤ |a| + |b|

x = x + y - y

|-y| = |y|

3. The attempt at a solution

|x + y| ≥ |x| - |y|

x = x + y - y (don't know where to use it)

xy ≤ |xy| = |x| |y| ( I really don't know why I am taking these steps, I am pretty much following the proof of theorem 3 in this book)

2xy ≤ 2|x||y| ( does the |-y| = |y| come into play here to flip the inequality? )

(x+y)^2 = x^2 + 2xy + y^2 ≤ x^2 - 2|x||y| + y^2 = (|x| - |y|)^2

Last edited: Oct 23, 2013
2. Oct 23, 2013

### pasmith

Is it not obvious that x + y - y = x + (y - y) = x + 0 = x?

3. Oct 23, 2013

### tiny-tim

hi jimmyly!

x = x + (y - y)

now can you see why it's true?
add some brackets in a different place

4. Oct 23, 2013

### tiny-tim

hi jimmyly!
y - y = 0

x + (y - y) = x + 0

x + 0 = x ?

5. Oct 23, 2013

@mark-44: How was it false? I mean if you could clarify that would be great. Wasn't trying to spread false information or anything.

6. Oct 23, 2013

### jimmyly

I don't understand why I would write x + y ≥ x + y
how does the subtraction turn into addition? does this mean that x + y = |x| - |y|? I see that -(-x) = + x
but its |x| - |y| not |x| + |-y|
or is my thinking wrong?

7. Oct 23, 2013

### jimmyly

Thank you, I understand why x = x + y - y it is very obvious! I am confused because I have no idea what it has to do with the proof and where to apply it when proving that |x+y| >= |x| - |y|

8. Oct 23, 2013

### Staff: Mentor

Tornado Dragon's post was full of errors, so I deleted it. Right at the start he tells you to remember that |x|= x. This is not true in general, particularly if x happens to be negative.

No, it's not. You can't just ignore the absolute value signs, which Tornado Dragon seems to have done.

9. Oct 23, 2013

### tiny-tim

apply Theorem 3 to x = x + y - y (with brackets in a suitable place)

10. Oct 23, 2013

@jimmyly ignore my post I have made a few mistakes in it. I will get back to you on it once I have gone it over it again.

edit: @mark44&jimmlyl: I did make en eroneous error but I did not ignore the absolute signs like that I will redo it and post it at a later tim.e.

Last edited: Oct 23, 2013
11. Oct 23, 2013

### jimmyly

okay so here is what i'm doing right now

|x+y| >= |x| - |y|
with x = x + y - y
I got
|x + y| >= |x| + |y| - |y| - |y|
cancelling the |y|
|x + y| >= |x| + |y|
am I on the right track? :)

12. Oct 23, 2013

### jimmyly

understandable

no worries we all make mistakes!

13. Oct 23, 2013

### Dick

Take your theorem 3, |a+b|<=|a|+|b|. Put a=x+y and b=(-y). What happens?

14. Oct 23, 2013

### tiny-tim

no!!

Theorem 3 applies to 3 things

use brackets to reduce the number of things to 3 !!

then you can apply Theorem 3

15. Oct 23, 2013

### jimmyly

so from this I got

|a + b| <= |a| + |b|
|(x+y) + (-y)| <= |x+y| + |-y|
|x + y - y| <= |x+y| + |-y|
|x| - |y| <= |x + y|

16. Oct 23, 2013

### jimmyly

oh that just did it didn't it?!

17. Oct 23, 2013

### Dick

Sure did!

18. Oct 23, 2013

### jimmyly

Would this be classified as a direct proof? I'm trying to learn proofs on my own so this is a little bit confusing to me. Thanks everyone for helping me out!

19. Oct 23, 2013

### jimmyly

Wow that's amazing. Thanks everyone! You are all wonderful

20. Oct 23, 2013

### tiny-tim

jimmyly, you seem to be worrying that there's something special about proofs

there isn't

if it starts with the question and finishes with the answer, it's a proof!