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Triangle Inequality Proof

  • Thread starter SMA_01
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
22,097
3,280

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

[tex]|x|-|y|\leq |x-y| ~~~~~~~~~~~~~~(1)[/tex]

Now, you need to prove the other inequality

[tex]-|x-y|\leq |x| - |y|[/tex]

You have done the hard work already. All you need to do now is to switch around x and y in (1).
 
  • #3
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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?
 
  • #4
22,097
3,280
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

[tex]|y|-|x|\leq |y-x|[/tex]

What if you multiply both sides by -1?? (Watch out, the inequality will reverse direction!!)
 
  • #5
22,097
3,280
Sorry, the last step was correct!

So ignore my previous post.

What is [itex]|-(x-y)|[/itex]?? Can you eliminate the -?
 
  • #6
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I just edited my previous response, I think I got it (I hope)...
 
  • #7
218
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One quick question, is it necessary for me to assume lxl is greater than or equal to lyl like I did in the beginning?
 
  • #8
22,097
3,280
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 [itex]|x|-|y|\geq -|x-y|[/itex], but apart from that it's fine.
 
  • #9
22,097
3,280
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
218
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I did in the beginning, but I guess it wasn't necessary.

Thank you for all your help :)
 

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