# Triangle inequality proof

1. Mar 31, 2008

### chocolatelover

1. The problem statement, all variables and given/known data
Prove that |x+y|</|x|+|y|

2. Relevant equations

3. The attempt at a solution

Assume that x and y are real numbers.
|5+2|</|5|+|7|
7</7
|-5-2|</|-5|+|-2|
7</7

I know that it is true by testing different numbers, but I'm not sure how to prove it. Could someone please show me how or give me a hint?

Thank you very much

2. Mar 31, 2008

### sutupidmath

since -|x|<x<|x|, and -|y|<y<|y| we add these to get

-|x|-|y|<x+y<|x|+|y|=>-(|x|+|y|)<x+y<|x|+|y|,

now from the abs value properties it immediately yeilds to:

|x+y|<|x|+|y| read ''<" as greater or equal to.

Last edited: Mar 31, 2008
3. Mar 31, 2008

### sutupidmath

Or we could prove it this way also: to prove it it is also sufficient to prove that

$$|x+y|^{2}\leq (|x|+|y|)^{2} ???$$ so we have

$$|x+y|^{2}=(x+y)^{2}=x^{2}+2xy+y^{2}=|x|^{2}+2xy+|y|^{2}<|x|^{2}+2|x||y|+|y|^{2}=(|x|+|y|)^{2}$$

hence we are done.

4. Mar 31, 2008

### chocolatelover

Thank you very much

Regards

5. Mar 18, 2010

### vrdfx

Why do absolute value properties yield to |x+y|<|x|+|y| from -(|x|+|y|)<x+y<|x|+|y|? Could someone please explain this further?

6. Mar 18, 2010

### Gib Z

I don't think its valid in general to add two inequalities like that. But if you accept up to there, then to see the final step is just seeing that if -A < a < A, then we can say |a| < A.

7. Mar 23, 2010

### Gib Z

Correction and sincere apologies to sutupidmath, you CAN add inequalities as long as they are pointing the same same direction which in this case in true. And that goes to make an extremely simple proof!

8. Mar 24, 2010

### Testify

I think vrdfx wasn't asking how -|x|<x<|x| + -|y|<y<|y| = -|x|-|y|<x+y<|x|+|y| => -(|x|+|y|)<x+y<|x|+|y| , but how -(|x|+|y|)<x+y<|x|+|y| yields |x+y|<|x|+|y|.

Does -(|x|+|y|) = |x+y| or something? I don't see how it could...