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

[SMA_01]

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.

 

[micromass]

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).

 

[SMA_01]

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?

 

[micromass]

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!)

 

[micromass]

Sorry, the last step was correct!

So ignore my previous post.

What is $|-(x-y)|$?? Can you eliminate the -?

 

[SMA_01]

I just edited my previous response, I think I got it (I hope)...

 

[SMA_01]

One quick question, is it necessary for me to assume lxl is greater than or equal to lyl like I did in the beginning?

 

[micromass]

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.

 

[micromass]

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.

 

[SMA_01]

I did in the beginning, but I guess it wasn't necessary.

Thank you for all your help :)