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

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SUMMARY

The discussion focuses on proving the triangle inequality, specifically the statement llxl-lyll≤lx-yl. Participants utilize the properties of absolute values and the triangle inequality to derive the necessary inequalities. Key steps include proving |x|-|y|≤|x-y| and manipulating the inequalities to establish the proof. The final conclusion confirms that the assumption of lxl≥lyl is not necessary for the proof's validity.

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  • Understanding of absolute value properties
  • Familiarity with the triangle inequality theorem
  • Basic algebraic manipulation skills
  • Knowledge of inequalities and their properties
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  • Study the properties of absolute values in depth
  • Learn more about the triangle inequality theorem and its applications
  • Practice algebraic manipulation of inequalities
  • Explore advanced topics in real analysis related to inequalities
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Mathematics students, educators, and anyone interested in understanding the triangle inequality and its proofs in real analysis.

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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