Re: Inequality Questions.
From: Thomas Mautsch (mautsch_at_math.ethz.ch)
Date: 03/21/05
- Next message: Dave Seaman: "Re: Distinct linear orderings on Z"
- Previous message: Giuseppe Bilotta: "Re: Epistemology 201: The Science of Science"
- In reply to: dalthman: "Re: Inequality Questions."
- Next in thread: dalthman: "Re: Inequality Questions."
- Reply: dalthman: "Re: Inequality Questions."
- Messages sorted by: [ date ] [ thread ]
Date: 21 Mar 2005 19:19:03 +0100
In news:<882ebff2.0503130803.60f99b52@posting.google.com> schrieb dalthman:
>
> (1) For complex numbers, a, b, c, prove the following inequality
> |a-b|root(1+|c|^2) + |b-c|root(1+|a|^2) >= |c-a|root(1+|b|^2)
>
> (2) For complex numbers, z, w, prove the following inequality
> |z-w| >= root(1+|z|^2) - root(1+|w|^2)
>
> And I've already got a solution. That's not my homework!
> I'm not even a student.
(1) looks like a very interesting property to me:
If one looks at Euclidean 3-space as IR^3 = IC x IR,
and chooses points with coordinates
A:= (a,0), B:= (b,0), C:=(c,0), D:= (0,1),
then (1) is essentially equivalent to the statement
that the three products
|AB|*|CD|, |CB|*|AD|, |AC|*|BD|
satisfy the triangle,
which is a generalization of Ptolemaios' theorem:
For A,B,C,D points in the plane,
there holds:
AC*BD <= AB*CD +BC*AD
with equality iff A,B,C,D are concircular.
Now, Ptolemaios' theorem has a very short proof
that uses only the properties of complex numbers,
which can be generalized to yield the following proof
of your inequality (1):
The vectors (a,1), (b,1), (c,1) from the 2-dimensional
complex vector space |C + |C satisfy the
complex-linear relation:
(a-b) * (c,1) + (b-c) * (a,1) = (a-c) * (b,1)
Now apply the Euclidean norm
(x,y) |--> || (x,y) || := sqrt( |x|^2 + |y|^2 )
to this formula,
use the triangle inequality,
and the fact that
|| z * (x,y) || = |z| * ||(x,y)||,
and you get:
|| (a-b) * (c,1) + (b-c) * (a,1) || = || (a-c) * (b,1) ||
|| (a-b) * (c,1) || + || (b-c) * (a,1) || >= || (a-c) * (b,1) ||
|a-b| * || (c,1) || + |b-c| * || (a,1) || >= |a-c| * || (b,1) ||
q.e.d.
Your other inequality (2) is *much* easier to see than (1)
is in terms of the geometric interpretation above.
P.S.: Who invented this generalized Ptolemaios' inequality in three-space?
- Next message: Dave Seaman: "Re: Distinct linear orderings on Z"
- Previous message: Giuseppe Bilotta: "Re: Epistemology 201: The Science of Science"
- In reply to: dalthman: "Re: Inequality Questions."
- Next in thread: dalthman: "Re: Inequality Questions."
- Reply: dalthman: "Re: Inequality Questions."
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|