# An Inequality in Cyclic Quadrilateral IV

### Proof

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,

$\displaystyle S=\frac{1}{2}ad\sin A+\frac{1}{2}bc\sin A,$

Or,

$\displaystyle\sin A=\frac{2S}{ad+bc}.$

Let $f:\;(0,2\pi)\rightarrow\mathbb{R},\;$

$f(x)=\sin x+\cos x =\sqrt{2}\cos\left(x+\frac{\pi}{4}\right).\;$

Thus $\max f(x)=\sqrt{2}.\;$ We now have $\displaystyle\sin x+\cos x\le\sqrt{2},\;$ i.e.

$\displaystyle \frac{a^2-b^2-c^2+d^2}{2(ad+bc)}+\frac{2S}{ad+bc}\le\sqrt{2},$

which is

$\displaystyle a^2-b^2-c^2+d^2+4S\le 2\sqrt{2}(ad+bc).$

### Acknowledgment

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