math 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 statment is correct.

thanks for your help i completely forgot those inequality rules :smile:

[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 im allowed to use it but in any case i know how to prove it thank you guys :smile:

[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|.