Proving the Triangle Inequality: How to Show llxl-lyll≤lx-yl

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Homework Help Overview

The discussion revolves around proving the triangle inequality, specifically the inequality llxl-lyll≤lx-yl. Participants are exploring the properties of absolute values and their relationships in the context of this inequality.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various approaches to proving the inequality, including assumptions about the magnitudes of x and y. There are attempts to manipulate the expressions using known properties of absolute values and the triangle inequality.

Discussion Status

Some participants have provided guidance on how to proceed with the proof, while others are questioning the necessity of certain assumptions made at the beginning of the discussion. Multiple interpretations of the steps involved are being explored.

Contextual Notes

There is a mention of a potential misunderstanding regarding the necessity of assuming lxl is greater than or equal to lyl, indicating that the participants are navigating through the assumptions that underpin the proof.

SMA_01
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Homework Statement



Prove llxl-lyll≤lx-yl

(The triangle inequality: la+bl≤lal+lbl)

The Attempt at a Solution



For the first part, I assumed lxl≥lyl:

lxl=l(x-y)+yl

Then, by Triangle Inequality

l(x+y)+yl≤l(x-y)l+lyl

So,
lxl≤l(x-y)l+lyl
Subtract lyl from both sides to get:

lxl-lyl≤l(x-y)l.

I'm not sure where to go from here. For, llxl-lyll≤lx-yl, don't I need to prove -l(x-y)l≤lxl-lyl≤l(x-y)l? How would I finish the proof?

Thank you.
 
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SMA_01 said:

Homework Statement



Prove llxl-lyll≤lx-yl

(The triangle inequality: la+bl≤lal+lbl)

The Attempt at a Solution



For the first part, I assumed lxl≥lyl:

lxl=l(x-y)+yl

Then, by Triangle Inequality

l(x+y)+yl≤l(x-y)l+lyl

So,
lxl≤l(x-y)l+lyl
Subtract lyl from both sides to get:

lxl-lyl≤l(x-y)l.

I'm not sure where to go from here. For, llxl-lyll≤lx-yl, don't I need to prove -l(x-y)l≤lxl-lyl≤l(x-y)l? How would I finish the proof?

Thank you.

Good. So you already proved

|x|-|y|\leq |x-y| ~~~~~~~~~~~~~~(1)

Now, you need to prove the other inequality

-|x-y|\leq |x| - |y|

You have done the hard work already. All you need to do now is to switch around x and y in (1).
 
micromass- Okay, I did that:

lyl=l(y-x)+xl

And then,

l(y-x)+xl≤l(y-x)l+lxl

so, lyl≤l(y-x)l+lxl

= lyl-lxl≤l(y-x)l

What do I do from here? Would I factor out a negative from both sides?

Like this:

-(lxl-lyl)≤l-(x-y)l

-(lxl-lyl)≤l-1ll(x-y)l

And since l-1l=1

-(lxl-lyl)≤l(x-y)l

Finally,
lxl-lyl≥l(x-y)l.

Is this correct?
 
SMA_01 said:
micromass- Okay, I did that:

lyl=l(y-x)+xl

And then,

l(y-x)+xl≤l(y-x)l+lxl

so, lyl≤l(y-x)l+lxl

= lyl-lxl≤l(y-x)l

What do I do from here? Would I factor out a negative from both sides?

Like this:

-(lxl-lyl)≤l-(x-y)l?

No, that last step is incorrect. You have

|y|-|x|\leq |y-x|

What if you multiply both sides by -1?? (Watch out, the inequality will reverse direction!)
 
Sorry, the last step was correct!

So ignore my previous post.

What is |-(x-y)|?? Can you eliminate the -?
 
I just edited my previous response, I think I got it (I hope)...
 
One quick question, is it necessary for me to assume lxl is greater than or equal to lyl like I did in the beginning?
 
SMA_01 said:
micromass- Okay, I did that:

lyl=l(y-x)+xl

And then,

l(y-x)+xl≤l(y-x)l+lxl

so, lyl≤l(y-x)l+lxl

= lyl-lxl≤l(y-x)l

What do I do from here? Would I factor out a negative from both sides?

Like this:

-(lxl-lyl)≤l-(x-y)l

-(lxl-lyl)≤l-1ll(x-y)l

And since l-1l=1

-(lxl-lyl)≤l(x-y)l

Finally,
lxl-lyl≥l(x-y)l.

Is this correct?

That last step should read |x|-|y|\geq -|x-y|, but apart from that it's fine.
 
SMA_01 said:
One quick question, is it necessary for me to assume lxl is greater than or equal to lyl like I did in the beginning?

Where do you assume that?? It doesn't seem necessary.
 
  • #10
I did in the beginning, but I guess it wasn't necessary.

Thank you for all your help :)
 

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