Inequality and absolute value proof

[EV33]

Homework Statement

prove that llal-lbll\leqla-bl

Homework Equations

Triangle inequality
lx+yl\leqlxl+lyl

The Attempt at a Solution

Let a=(a-b)+b
By using the triangle inequality we get
lal-lbl\leqla-bl

Then from here I am not sure what I can do. I would like to say on the left hand side that I can use the triangle inequality again by taking the absolute value of the left side and saying that I have the absolute value of a plus -absolute value of b.

llal+(-lbl)l\leqlal-lbl\leqla-bl

My second thought it instead of doing that I could possible just take the absolute value of both sides and then I would get what I was trying to prove because taking the absolute value of the right side wouldn't change anything, but taking the absolute value of the left side would change it exactly the way I need it to be.

I was wondering if either one of those Ideas is legitament because I am really not sure about them. If neither one is could someone point me in the right direction in proving this. Thank you for your time.

 

[EV33]

Sorry I will have to edit this. I don't know what happened.

 

[EV33]

fixed it

 

[Inferior89]

In general taking absolute value of both sides does not preserve an inequality.

For example:
-2 < -1
|-2| > |-1|

So that doesn't work.

Your first method seems to work much better.

 

[EV33]

So the first method works for sure? It seems reasonable to me but someone I talked to seemed like they thought it might not work. So I just want to make sure.

Thank you.

 

[HallsofIvy]

Yes, your first proof works. However, you shouldn't say "let a= (a- b)+ b". You don't need to "let" that be true- it is! What you mean to say is "in |x+ y|\le |x|+ |y|, let x= a- b and y= b". Then x+ y= (a- b)+ b= a so |a|\le |a- b|+ |b| and, subtracting |b| from both sides, |a|- |b|\le |a- b|.