Wonderful Inequality on Unit Circle

Problem

Wonderful Inequality on Unit Circle

Proof 1

Let's set $a=e^{2ix},\,b=e^{2iy},\,$ $x,y\in\mathbb{R}.\,$ Then

$\displaystyle\begin{align} \left(\frac{a+b}{1+ab}\right)^2 &= \left(\frac{(\cos 2x+\cos 2y)+i(\sin 2x+\sin 2y)}{1+\cos (2x+2y)+i\sin (2x+2y)}\right)^2\\ &= \left(\frac{\cos (x-y)(\cos (x+y)+i\sin(x+y))}{\cos (x+y)(\cos (x+y)+i\sin (x+y))}\right)^2\\ &=\frac{\cos ^2(x-y)}{\cos^2(x+y)}. \end{align}$

Similarly, $\displaystyle \left(\frac{a-b}{1-ab}\right)^2=\frac{\sin ^2(x-y)}{\sin^2(x+y)}.\,$ Adding up:

$\displaystyle\begin{align} \left(\frac{a+b}{1+ab}\right)^2 +\left(\frac{a-b}{1-ab}\right)^2&= \frac{\cos ^2(x-y)}{\cos^2(x+y)}+\frac{\sin ^2(x-y)}{\sin^2(x+y)}\\ &\ge \frac{\cos ^2(x-y)+\sin^2(x-y)}{\max\{\cos^2(x+y),\sin^2(x+y)\}}\\ &=\frac{1}{\max\{\cos^2(x+y),\sin^2(x+y)\}}\\ &= 1. \end{align}$

Proof 2

Introduce $x^2=a\,$ and $y^2=b,\,$ $|x|=|y|=1.\,$ We have

$\displaystyle\left(\frac{a+b}{1+ab}\right)^2 = \frac{\displaystyle\left(\frac{x}{y}+\frac{y}{x}\right)^2}{\displaystyle \left(xy+\frac{1}{xy}\right)^2}\;$ and $\displaystyle\left(\frac{a-b}{1-ab}\right)^2 = \frac{-\displaystyle\left(\frac{x}{y}-\frac{y}{x}\right)^2}{-\displaystyle \left(xy-\frac{1}{xy}\right)^2}.\;$

Further, $\displaystyle 2=|xy|+\frac{1}{|xy|}\ge\left| xy+\frac{1}{xy}\right|,\,$ implying $\displaystyle 4\ge\left| xy+\frac{1}{xy}\right|^2=\left(xy+\frac{1}{xy}\right)^2,\,$ because $\displaystyle xy+\frac{1}{xy}\,$ is real. We conclude that $\frac{\displaystyle\left(\frac{x}{y}+\frac{y}{x}\right)^2}{\displaystyle \left(xy+\frac{1}{xy}\right)^2}\ge\frac{\displaystyle\left(\frac{x}{y}+\frac{y}{x}\right)^2}{\displaystyle 4}.\;$ Similarly, $\displaystyle 4\ge\left| xy-\frac{1}{xy}\right|^2=-\left(xy-\frac{1}{xy}\right)^2,\,$ because $\displaystyle xy-\frac{1}{xy}\,$ is purely imaginary. We conclude that $\frac{-\displaystyle\left(\frac{x}{y}-\frac{y}{x}\right)^2}{\displaystyle -\left(xy-\frac{1}{xy}\right)^2}\ge \frac{-\displaystyle\left(\frac{x}{y}-\frac{y}{x}\right)^2}{\displaystyle 4}.\;$

It follows that

$\displaystyle\left(\frac{a+b}{1+ab}\right)^2+\left(\frac{a-b}{1-ab}\right)^2\ge\frac{\displaystyle\left(\frac{x}{y}+\frac{y}{x}\right)^2}{4}+\frac{-\displaystyle\left(\frac{x}{y}-\frac{y}{x}\right)^2}{4}\ge 1.$

Proof 3

We will prove a generalization:

$\displaystyle\left(\frac{a+b}{r^2+ab}\right)^2+\left(\frac{a-b}{r^2-ab}\right)^2\ge \frac{1}{r^2},$

for $|a|=|b|=r\gt 0.\,$ Our proposed generalization is equivalent to:

$\displaystyle\left(\frac{r(a+b)}{r^2+ab}\right)^2+\left(\frac{r(a-b)}{r^2-ab}\right)^2\ge 1.$

Letting $a=re^{ix}\,$ and $b=re^{iy},\,$ $r\gt 0,\,$ we have

$\displaystyle\begin{align} \frac{r(a+b)}{r^2+ab} &= \frac{r^2\left(e^{ix}+e^{iy}\right)}{r^2\left(1+e^{i(x+y}\right)}\\ &= \frac{e^{i(x+y)/2}\left(e^{i(x-y)/2}+e^{-i(x-y)/2}\right)}{e^{i(x+y)/2}\left(e^{i(x+y)/2}+e^{-i(x+y)/2}\right)}\\ &=\frac{\cos (x-y)}{\cos (x+y)}. \end{align}$

Similarly, $\displaystyle\frac{r(a-b)}{r^2-ab}=\frac{\sin (x-y)}{\sin (x+y)}.\,$ Therefore,

$\displaystyle\begin{align} \left(\frac{r(a+b)}{r^2+ab}\right)^2+\left(\frac{r(a-b)}{r^2-ab}\right)^2 &= \frac{\cos^2 (x-y)}{\cos^2 (x+y)}+\frac{\sin^2 (x-y)}{\sin^2 (x+y)}\\ &\ge \cos^2(x-y)+\sin^2(x-y)=1. \end{align}$

Setting $r=1\,$ yields the original proposed inequality.

Proof 4

The first (second) unsquared fraction is real as numerator and denominator lie on the inner angle bisector of $a\,$ and $b\,$ ($-b\,).$ Suppose that the first squared fraction is less than $1:\,$ the angle $\varphi\,$ between $a\,$ and $b,\,$ which is taken in $[0,\,\pi],\,$ is then larger than the angle $\psi\,$ between $ab\,$ and $1.\,$ But the angle between $a\,$ and $-b\,$ is $\pi-\varphi\,$ and the angle between $-ab\,$ and $1\,$ is the larger angle $\pi-\psi,\,$ hence the second squared fraction is then greater than $1$.

Proof 5

For $a=\cos\alpha +i\sin\alpha\,$ and $b=\cos\beta +i\sin\beta\,$

$\displaystyle\begin{align} a+b &= 2\cos\frac{\alpha-\beta}{2}\left(\cos\frac{\alpha+\beta}{2}+i\sin\frac{\alpha+\beta}{2}\right)\\ 1+ab &= 2\cos\frac{\alpha+\beta}{2}\left(\cos\frac{\alpha+\beta}{2}+i\sin\frac{\alpha+\beta}{2}\right)\\ a-b &= -2\sin\frac{\alpha-\beta}{2}\left(\sin\frac{\alpha+\beta}{2}-i\cos\frac{\alpha+\beta}{2}\right)\\ 1-ab &= -2\sin\frac{\alpha+\beta}{2}\left(\sin\frac{\alpha+\beta}{2}-i\cos\frac{\alpha+\beta}{2}\right). \end{align}$

It follows that

$\displaystyle\frac{\displaystyle\cos^2\frac{\alpha-\beta}{2}}{\displaystyle\cos^2\frac{\alpha+\beta}{2}}+\frac{\displaystyle\sin^2\frac{\alpha-\beta}{2}}{\displaystyle\sin^2\frac{\alpha+\beta}{2}}\ge 1$

is equivalent to

$\begin{align} (1+\cos(\alpha-\beta))(1-\cos(\alpha+\beta))&+(1-cos(\alpha-\beta))(1+\cos(\alpha+\beta))\\ &\ge 1-\cos^2(\alpha+\beta), \end{align}$

or,

$\cos^2(\alpha+\beta)-2\cos(\alpha-\beta)\cos(\alpha+\beta) +1\ge 0.$

The discriminant of the quadratic (in $\cos(\alpha-\beta))$ expression on the left equals $\cos^2(\alpha-\beta)-1\le 0,\,$ implying that that expression never changes its sign and, thus, proving the inequality.

Acknowledgment

The problem has been posted at the CutTheKnotMath facebook page by Leo Giugiuc (Romania) who credited it to Qing Song (China); the two solutions above are by Leo Giugiuc (Romania). After the page has been announced on facebook, Henry Ricardo (NY, NY) has informed us that the problem appeared as #757 in The Pentagon journal. The problem has been proposed by Jose Luis Diaz-Barrero (Spain) and Henry Ricardo's solution (Proof 3) has been published in the Fall 2015 issue of the magazine. Proof 4 is by Grégoire Nicollier (Switzerland). Leo Giugiuc has found a solution (Proof 5) by Stanciu Neculai (Romania).

 

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

70554180