Triangle ineuality, in need for help.

[MathematicalPhysicist]

i need help in proving this essential ineuality which i don't know how to prove (quite trivial isn't it):
||x|-|y||<=|x-y|

i know that my first trick is this:
|x|=|(x-y)+y|<=|(x-y)|+|y|
|x|-|y|<=|x-y|
and iv'e been told to do so also with |y| and i get to this:
|y|-|x|<=|y-x|
and then adding both inequalities give us with this:
|x-y|>=-|y-x|

here I'm pretty much puzzled, so can you help, or add some insight or hints.

btw, I'm in a hurry so if i have angried anyone by posting here, sorry!

 

[devious_]

You were almost there! You just needed this little step:
|y|-|x| <= |y-x|
=> |x|-|y| >= -|y-x| = -|x-y|

So you have -|x-y|<=|x|-|y|<=|x-y|.

 

[mathman]

Another way of seeing it is that ||x|-|y||=|x|-|y| or =|y|-|x|, depending on which is bigger , |x| or |y|.

 

[Galileo]

|x|-|y|<=|x-y|
and iv'e been told to do so also with |y| and i get to this:
|y|-|x|<=|y-x|

You don't have to add the inequalities. You've shown that for any x and y that |x-y| is greater-equal to both |y|-|x| and |x|-|y|. So isn't it obvious it's also greater-equal to the absolute value of |y|-|x|?

 

[MathematicalPhysicist]

You don't have to add the inequalities. You've shown that for any x and y that |x-y| is greater-equal to both |y|-|x| and |x|-|y|. So isn't it obvious it's also greater-equal to the absolute value of |y|-|x|?

apparently no, you need to prove up until you can say q.e.d or מ.ש.ל

 

[MathematicalPhysicist]

You were almost there! You just needed this little step:
|y|-|x| <= |y-x|
=> |x|-|y| >= -|y-x| = -|x-y|
So you have -|x-y|<=|x|-|y|<=|x-y|.

i think you also jumped too quickly here, don't you need to divise the options when |y-x|= y-x when y-x>=0 and |y-x|=x-y when y-x<0 and then you can see that -|y-x|=-|y-x|.

ok, i get it.
thank you.
a great site we have here, please don't change physicsforums. (-:

 

[arildno]

apparently no, you need to prove up until you can say q.e.d or מ.ש.ל

Remember that ||x|-|y||=max(|x|-|y|,|y|-|x|)
Reconsider Galileo's post.

 

[MathematicalPhysicist]

well, arlidno i think your notation is with regard to vector calclulus which iv'e seen your notation from, but my question is about vector calclulus and this is why i don't think it's related here, but you are the expert here not me.

cheers.

 

[Galileo]

So are x and y vectors or numbers? (Note: It doesn't matter!)

Ok, if you have some number a that is larger than b and larger than -b, you don't think it's obvious it's also larger than |b|?

Same thing here: You have some number a=|x-y| larger than b=|x|-|y| and larger than -b=|y|-|x| (or equal ofcourse). So a>=|b| or |x-y|>=||x|-|y||

Also, reread arildno's post. Clearly |a| is either a or -a, whichever is positive and thus the mximum max(a,-a) of a and -a.

 

[MathematicalPhysicist]

So are x and y vectors or numbers? (Note: It doesn't matter!)
Ok, if you have some number a that is larger than b and larger than -b, you don't think it's obvious it's also larger than |b|?
Same thing here: You have some number a=|x-y| larger than b=|x|-|y| and larger than -b=|y|-|x| (or equal ofcourse). So a>=|b| or |x-y|>=||x|-|y||
Also, reread arildno's post. Clearly |a| is either a or -a, whichever is positive and thus the mximum max(a,-a) of a and -a.

ok i understand your reasoning, but then again this question if we would be on topic was a help on pure mathematical question, and the real question is:
will the checker approve this obviously logical assertion, this i don't know.

but it looks that your way is the same as of devious and mathman has proposed, i myself not yet convinced it's any different, just other notation.

but I'm willing to be disproven.

 

[rachmaninoff]

Or you can just square both sides:

$$||x|-|y||^2=(|x|-|y|)^2=x^2-2|x||y|+y^2$$
$$|x-y|^2=(x-y)^2=x^2-2x\cdot y+y^2$$

where of course, multiplication is the scalar dot product in Euclidean space, or more generally the inner product.

 

[leon1127]

cant you just square both sides?