Math Triangle-inequality problem

[Anzas]

prove that for any real values of x,y
| |x|-|y| | <= |x+y| <= |x| + |y|

 

[marlon]

Well try using these properties :

1) |x| <= a then -a <= x <= a
2) in order to loose the ||-signs make a distinction for |x| when x > 0 and x < 0
If x > 0 then |x| = x
if x < 0 then |x| = -x
3) Triangle-inequality : |x| - |y| <= |x + y| <= |x| + |y|

I am not saying you will need all this properties, only some of them will do to make your proof. How would you start ? You have all the necessary "ingredients" when it comes to the ||-properties

regards
marlon
good luck

 

[Anzas]

ok i think i got it now
| |x|-|y| | <= |x+y| <= |x|+|y|

ill mark |x+y| as "a"

| |x|-|y| | <= a
then
|x|-|y| <= a
which gives
|x|-|y| <= |x+y| <= |x|+|y|
and by triangle inequality we can see that this statement is correct.

thanks for your help i completely forgot those inequality rules

 

[Gokul43201]

The question is basically asking you to prove the triangle inequality.

I doubt that you'd be allowed to used the triangle inequality to prove the same.

You can prove it either by considering the 4 cases where x and y are positive and negative reals, or using trigonometry (cosine rule) to prove it in the general case of complex numbers, which would automatically make it true in the reals.

 

[marlon]

Good point by Gokul...

Can you use the triangle identity yes or no? If not, there is a nice way to prove the triangle-identity but it is usually given as theory...there different proofs of different levels using different "kinds" of math

marlon

 

[Anzas]

im pretty sure I am allowed to use it but in any case i know how to prove it thank you guys

 

[Hurkyl]

| |x|-|y| | <= a
then
|x|-|y| <= a
which gives
|x|-|y| <= |x+y| <= |x|+|y|

You've done things backwards! You used | |x|-|y| | <= a to prove |x|-|y| <= |x+y|, but your goal was to prove | |x|-|y| | <= a, not |x|-|y| <= |x+y|.