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, dFurthermore 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.