# An Inequality in Cyclic Quadrilateral III

### Proof 1

In a cyclic quadrilateral, $A+C=B+D=180^{\circ},\;$ implying $\displaystyle\frac{C}{2}=90^{\circ}-\frac{A}{2}\;$ and $\displaystyle\frac{D}{2}=90^{\circ}-\frac{B}{2}.\;$ It follows that

\displaystyle\begin{align} \cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\cos\frac{D}{2} &= \cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{A}{2}\sin\frac{B}{2}\\ &=\frac{1}{4}\sin A\sin B. \end{align}

We'll now refer to a previous statement, viz.,

\displaystyle\begin{align} \sin A\sin B &\le (\frac{s}{a}-1)(\frac{s}{b}-1)(\frac{s}{c}-1)(\frac{s}{d}-1)\\ &=\frac{(s-a)(s-b)(s-c)(s-d)}{abcd}\\ &=\frac{S^2}{abcd}, \end{align}

by Brahmagupta's theorem. The combination of the two proves the required inequality.

### Proof 2

In $\Delta ABD: BD^2=a^2+d^2-2ad\cos A,\;$

In $\Delta BCD: BD^2=b^2+c^2-2bc\cos (\pi-A).$

It follows that

$a^2+d^2-2ad\cos A=b^2+c^2+2bc\cos A,$

or,

(1)

$\displaystyle\cos A=\frac{a^2-b^2-c^2+d^2}{2(ad+bc)}.$

Further,

\displaystyle\begin{align} \cos^2 \frac{A}{2}&=\frac{1+\cos A}{2}=\frac{1+\displaystyle\frac{a^2-b^2-c^2+d^2}{2(ad+bc)}}{2}\\ &=\frac{2ad+2bc+a^2-b^2-c^2+d^2}{4(ad+bc)}=\frac{(a+d)^2-(b-c)^2}{4(ad+bc)}\\ &=\frac{(a+d-b+c)(a+d+b-c)}{4(ad+bc)}=\frac{(2s-2b)(2s-2c)}{4(ad+bc)}. \end{align}

So that $\displaystyle\cos^2 \frac{A}{2}=\frac{(s-b)(s-c)}{ad+bc}.\;$ Simialrly,

$\displaystyle\cos^2 \frac{C}{2}=\frac{(s-a)(s-d)}{ad+bc},\;$ etc.

Finally,

\displaystyle\begin{align} \cos^2 \frac{A}{2}\cos^2 \frac{B}{2}\cos^2 \frac{C}{2}\cos^2 \frac{D}{2}&=\frac{(s-a)^2(s-b)^2(s-c)^2(s-d)^2}{(ad+bc)^2(ab+cd)^2}\\ &=\Biggl(\frac{S^2}{(ad+bc)(ab+cd)}\Biggl)^2\\ &\leq \Biggl(\frac{S^2}{2\sqrt{abcd}\cdot 2\sqrt{abcd}}\Biggl)^2 \end{align}

And the result follows:

$\displaystyle\displaystyle \cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} \cos \frac{D}{2}\leq \frac{S^2}{4abcd}$

### Acknowledgment

The problem from his book Math Accent has been posted at the CutTheKnotMath facebook page by Dan Sitaru, with his solution (Proof 2).