Is There a Way to Prove the Triangle Inequality for Absolute Values?

[chocolatelover]

Homework Statement

Prove that |x+y|</|x|+|y|

Homework Equations

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

 

[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.

 

[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.

 

[chocolatelover]

Thank you very much

Regards

 

[vrdfx]

-(|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.

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

 

[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.

 

[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!

 

[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...