# 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).