I come across the following problem in my study.

Consider in the real field. Let $ 0\le x\le1 $, $a_1^2+a_2^2=b_1^2+b_2^2=1$.Is it true

$ (a_1b_1+xa_2b_2)^2\le\left(\frac{(1-x)+(1+x)(a_1b_1+a_2b_2)}{(1+x)+(1-x)(a_1b_1+a_2b_2)}\right)^{2}(a_1^2+xa_{2}^{2})(b_1^2+xb_{2}^{2})$?

  • 5
    $\begingroup$ Something is wrong with your inequality: when I substitute $x=0$ it implies $a_1b_1\le(a_1b_1)^2$, while the opposite inequality holds. $\endgroup$ – Wadim Zudilin Jun 1 '10 at 4:56
  • $\begingroup$ Hi miwalin, I think you need, modify the inequality a little bit. Because if we take $x=1$, we would get that $(a_1b_1+a_2b_2)\leq (a_1b_1+a_2b_2)^2.$ But choosing vectors $(a_1,a_2)$ and $(b_1,b_2)$ in the unit circle, such that the angle between them is close to $\frac{\pi}{2}$, we see that the inequality is not hold in general. $\endgroup$ – Leandro Jun 1 '10 at 4:59
  • $\begingroup$ I modified it. Thank you for pointing out my typos. $\endgroup$ – Sunni Jun 1 '10 at 6:03
  • $\begingroup$ Your older question has the exact same title: mathoverflow.net/questions/20172/a-plausible-inequality. This will make it difficult for people to browse questions in the future. $\endgroup$ – j.c. Jun 1 '10 at 7:15
  • $\begingroup$ title changed from "A Plausible Inequality" $\endgroup$ – Gerald Edgar Jun 1 '10 at 12:30

As Can Hang points out in his response, the inequality does not hold in general. Thanks to his comment to my own post, I stand corrected and claim the inequality is valid at least for the case $$ a_1b_1+xa_2b_2\ge0 \qquad(*) $$ (and this seems to be a necessary condition as well).

Let me do some standard things. First let $$ a_1=\frac{1-u^2}{1+u^2}, \quad a_2=\frac{2u}{1+u^2}, \quad b_1=\frac{1-v^2}{1+v^2}, \quad b_2=\frac{2v}{1+v^2} $$ where $uv\ge0$. Substitution reduces the inequality to the following one: $$ ((1-u^2)(1-v^2)+4xuv)^2 \le\biggl(\frac{(uv+1)^2-x(u-v)^2}{(uv+1)^2+x(u-v)^2}\biggr)^2 ((1-u^2)^2+4xu^2)((1-v^2)^2+4xv^2). \qquad{(1)} $$ Now introduce the notation $$ A=(1-u^2)(1-v^2)+4xuv, \quad B=(uv+1)^2, \quad C=x(u-v)^2 $$ and note that $A,B,C$ are nonnegative; the inequality $A\ge0$ is equivalent to the above condition $(*)$. In addition, $$ A\le B-C \qquad{(2)} $$ because $$ B-C-A=(1-x)(u+v)^2\ge0. $$ In the new notation the inequality (1) can be written more compact: $$ A^2(B+C)^2\le(B-C)^2(A^2+4BC) $$ which after straightforward reduction becomes $$ A^2\le(B-C)^2, $$ while the latter follows from (2).

  • $\begingroup$ $A$ is not necessarily nonnegative. For example, $x=1,u=1,v=-1$. $\endgroup$ – Sunni Jun 1 '10 at 15:58
  • $\begingroup$ Oh-oh, you were very careful in checking. I'll fix when I have time. But why do you expect from me a cleaned solution?! $\endgroup$ – Wadim Zudilin Jun 2 '10 at 0:58
  • $\begingroup$ I found, by accidently, that your proof relies on the nonnegativity of $A$. I just point out for you. $\endgroup$ – Sunni Jun 2 '10 at 1:24
  • $\begingroup$ You know now that the inequality does not work in general, but a tiny (natural! see the second post) condition insures that it holds. $\endgroup$ – Wadim Zudilin Jun 2 '10 at 1:58
  • 1
    $\begingroup$ The condition is probably not too strange: your inequality is for the scalar product of two vectors, $(a_1,\sqrt{x}a_2)$ and $(b_1,\sqrt{x}b_2)$. The angle between them should be cute. I am still curious about some "geometric" proof of the inequality. $\endgroup$ – Wadim Zudilin Jun 2 '10 at 5:02

I think your inequality is false, dear miwalin. Please check the case when $a_1=b_2=\frac{\sqrt{3}}{2}$ and $a_2=b_1=-\frac{1}{2}.$ But I think it is true when $a_1,$ $a_2,$ $b_1,$ $b_2$ are nonnegative numbers.

Let me prove it in the case $a_1,$ $a_2,$ $b_1,$ $b_2$ are nonnegative real numbers. Write the inequality as $$\frac{(a_1^2+xa_2^2)(b_1^2+xb_2^2)}{(a_1b_1+xa_2b_2)^2} -1 \ge \left[ \frac{(1+x)+(1-x)(a_1b_1+a_2b_2)}{(1-x)+(1+x)(a_1b_1+a_2b_2)}\right]^2-1.$$ Since $$(a_1^2+xa_2^2)(b_1^2+xb_2^2)-(a_1b_1+xa_2b_2)^2=x(a_1^2b_2^2+a_2^2b_1^2-2a_1a_2b_1b_2)= x[(a_1^2+a_2^2)(b_1^2+b_2^2)-(a_1b_1+a_2b_2)^2]= x[1-(a_1b_1+a_2b_2)^2]$$ and $$\left[ \frac{(1+x)+(1-x)(a_1b_1+a_2b_2)}{(1-x)+(1+x)(a_1b_1+a_2b_2)}\right]^2-1=\frac{4x[1-(a_1b_1+a_2b_2)^2]}{[(1-x)+(1+x)(a_1b_1+a_2b_2)]^2},$$ the above inequality is equivalent to (notice that $x[1-(a_1b_1+a_2b_2)^2] \ge 0$) $$[(1-x)+(1+x)(a_1b_1+a_2b_2)]^2 \ge 4(a_1b_1+xa_2b_2)^2,$$ or $$(1-x)+(1+x)(a_1b_1+a_2b_2) \ge 2(a_1b_1+xa_2b_2),$$ or $$(1-x)(1-a_1b_1+a_2b_2) \ge 0,$$ which is obvious.

  • $\begingroup$ What you take the value of $x$? $\endgroup$ – Sunni Jun 2 '10 at 1:24
  • $\begingroup$ After taking my values of $a_1,$ $a_2,$ $b_1,$ $b_2,$ the inequality is false for all $0<x<1.$ $\endgroup$ – can_hang2007 Jun 2 '10 at 1:26
  • $\begingroup$ @Can Hang: +1. Thanks for figuring out an explicit counter example. The inequality is true for $a_2b_2>0$. $\endgroup$ – Wadim Zudilin Jun 2 '10 at 2:23
  • $\begingroup$ Well, your assistance is extremely helpful. Thank you! The correct condition is $a_1b_1+xa_2b_2\ge0$. $\endgroup$ – Wadim Zudilin Jun 2 '10 at 3:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.