# Another plausible inequality.

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

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

• 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. – Wadim Zudilin Jun 1 '10 at 4:56
• 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. – Leandro Jun 1 '10 at 4:59
• I modified it. Thank you for pointing out my typos. – Sunni Jun 1 '10 at 6:03
• 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. – j.c. Jun 1 '10 at 7:15
• title changed from "A Plausible Inequality" – 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).

• $A$ is not necessarily nonnegative. For example, $x=1,u=1,v=-1$. – Sunni Jun 1 '10 at 15:58
• 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?! – Wadim Zudilin Jun 2 '10 at 0:58
• I found, by accidently, that your proof relies on the nonnegativity of $A$. I just point out for you. – Sunni Jun 2 '10 at 1:24
• You know now that the inequality does not work in general, but a tiny (natural! see the second post) condition insures that it holds. – Wadim Zudilin Jun 2 '10 at 1:58
• 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. – 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.

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