Date: Thu, 27 Apr 95 13:35:09 CDT
From: rusin (Dave Rusin)
To: [Anonymous]
Subject: Re: Question: Does these two inequalities imply THIS INEQUALITY???

In article [deleted] wrote:
>I have this (possibly easy?) question: 
>      Let b be a real number and let a, c, d<e  anf f<e be positive numbers. 
>Furthermore suppose that  
>                        (ax^2 +bxy + cy^2) is less or equal  to (dx^2+ey^2)
>and                  (ax^2 +bxy + cy^2) is less or equal  to (ex^2+fy^2)
>Does this imply that 
>                        (ax^2 +bxy + cy^2) is less or equal  to  (dx^2+fy^2)? 

I thought about your problem but did not reach a solution. I did get as
far as eliminating  x  and  y, that is, describing it as a problem with
inequalties on  a, b, c, d, e, f. Perhaps this will be useful.
I assumed your quadratic form was positive definite, as you mentioned later
in your post.

Observe that an ellipse centered at the origin contains the unit circle
iff the semi-minor axis has length at least 1. If this ellipse is written
E={ v in R^2 : transpose(v).A.v = 1 }, then the semi-minor axis is the
reciprocals of the square root of the larger eigenvalue of  A.
So an ellipse given by  A x^2 + B xy + C y^2 = 1 contains the unit
circle iff the polynomial
	X^2 - (A+C) X + (AC-B^2/4) = 0
has (both roots positive and) the larger root less than  1. This is 
equivalent to the condition that  ( B^2 < 4 AC  and )  B^2 < 4 (A-1)(C-1)
and  A+C < 2.

On the other hand, the condition that the ellipse contain the circle is
equivalent to saying that   Ax^2+Bxy+Cy^2=1  ==>  x^2+y^2 >= 1, or 
in other words,  Ax^2+Bxy+Cy^2 <= x^2+y^2  for points on the ellipse.
Since both polynomials are homogeneous, this is equivalent to having the
inequality hold for all  x  and  y.

Next observe that a simple scaling of the variables  x  and  y  shows
	ax^2+bxy+cy^2 <= dx^2+ey^2  for all  x, y      iff
	(a/d)u^2 + (b/sqrt(de))uv + (c/e) v^2 <= u^2+v^2  for all  u,v.
By the previous two paragraphs, this holds iff
	(  b^2/de < 4 (a/d)(c/e)  and )
	b^2/de < 4 (a/d - 1)(c/e - 1)  and
	(a/d) + (c/e) < 2
which may be written as  (b^2<4ac and)
	b^2 < 4 (a-d)(c-e)   and  ae + dc < 2de

OK, so if I've done the algebra correctly, you're assuming (b^2<4ac and)
	b^2 < 4 (a-d)(c-e)   and  ae + cd < 2de
and	b^2 < 4 (a-e)(c-f)   and  af + ce < 2ef
and you want to know if that implies
	b^2 < 4 (a-d)(c-f)   and  af + cd < 2df
This expresses the original problem without the  x  and y.

I was hoping to complete this analysis but I think I'm running low on
ideas. Perhaps you will want to finish in one of several ways.
(You will want to keep track of the other inequalities you began
with -- I haven't used them yet.)

(1) Let  u1=-b^2 + 4(a-d)(c-e)  and so on. Then the first four inequalities
state that  u1,... are positive. So is  u0= 4ac-b^2. Use the equations 
defining  these u_i  to solve for (say) d, e, f, a, and  c  in terms of
b and  u0...u4. Then substitute into the last proposed inequalities and
see if you can recognize them as positive quantities.

(2) Perform a lagrange-multiplier analysis on the region defined by
the given five inequalities (u_i >= 0) and search for the minimum of
the last two expressions, checking to see if positive.

(3) Work geometrically: For any fixed a,b,c the sets of d,e,f that
satisfy these pairs of inequalities may be easily computed. The first
pair of inequalties, for example, describe the region bounded by two hyperbolas
	(d-a)(e-c) > b^2/4    and  (d-a/2)(e-c/2) > ac/4
Since  f  is not mentioned here, the solution set in  d,e,f-space
is bounded by hyperbolic sheets. Likewise the second pair describes another
such region. Your goal is to show the first regions' intersection includes
the region bounded by a third set of hyperbolic sheets.
	In this method of attack, notice that the hyperbola
(d-a/2)(e-c/2) = ac/4  passes through the origin in the d-e-plane, so since
you assumed in your post that  d, e > 0, the solution set to 
(d-a/2)(e-c/2) > ac/4 is just the single region  past d=a/2, e=c/2.
(The other hyperbola in the d-e-plans, unfortunately, seems to involve
two components.)

I didn't pursue your more general problem.

dave

==============================================================================

From: [Anonymous]
To: rusin@math.niu.edu (Dave Rusin)
Date:          Thu, 27 Apr 1995 22:03:01 MET-1
Subject:       Re: Question: Does these two inequalities imply THIS INEQUA

Dear Dave.  I am really thankful for your effort.
  Actually, I am hoping that there excist a counter-example to this 
statement.  You see, I have discovered an estimate concerning non-linear 
partial differential operators which for the special case when these 
operators are linear, seems to be better than some other well known 
estimates. These well known estimates (inequalities) are in R^2 of the same form
 as those two inequalities  in my question. What they  (hopefully 
not) imply is my estimate (inequality). 
  If you find out something more about my question, pleace contact 
me!
  I am looking forward for your reply.


==============================================================================

Date: Fri, 28 Apr 95 00:02:16 CDT
From: rusin (Dave Rusin)
To: rusin
Subject: counterexample

a=c=2, b=3; d=f=3, e=9.

In the earlier letter I should have noted that we have
	(b^2/4ac) = (d/a - 1)(e/c - 1)
so the factors on the right have the same sign. If both were negative,
a/d >1 and c/e > 1 which is prohibited by a/d+c/e < 2. Thus both are
positive, andthe other inequality is unnecessary.

So we have an  R  between  0 and  1  and know
	R < ( d/a - 1 ) (e/c - 1) ,  d/a > 1,  e/c > 1.
	R < ( e/a - 1 )( f/c - 1) ,  e/a > 1,  f/c > 1.
Thus  d/a > 1  and  f/c > 1  are automatic. The question is whether
(d/a -1)(f/c-1)  is still greater than  R.  This seems unlikely if
e  is much larger than  d or f, since then we could get a and c just
less than  d  and  f  respectively, and make the two inequalities true
without the last.