# An Inequality in Cyclic Quadrilateral II

### Proof

By Brahmagupta's theorem, the required inequality is equivalent to

$abcd\sin A\sin B\le [ABCD]^2,$

where $[ABCD]\;$ denotes the area of the quadrilateral. By the AM-GM inequality, this is equivalent to

$abcd\sin A\sin B\le \displaystyle\frac{(ad+bc)\sin A}{2}\cdot\frac{(ab+cd)\sin B}{2},$

meaning $abcd\le \displaystyle\frac{(ad+bc)}{2}\cdot\frac{(ab+cd)}{2}.$

But, by the AM-GM inequality, $\displaystyle\sqrt{abcd}\le\frac{ad+bc}{2}\;$ and, similarly, $\displaystyle\sqrt{abcd}\le\frac{ab+cd}{2}.\;$ The product of the two is the required $abcd\le \displaystyle\frac{(ad+bc)}{2}\frac{(ab+cd)}{2}.$

Equality is achieved when $ad=bc\;$ and $ab=cd.\;$ Taking the product: $a^2bd=c^2bd,\;$ or $a^2=c^2,\;$ and, subsequently, $a=c.\;$ But then $b=d,\;$ implying that $ABCD\;$ is a parallelogram, and, being cyclic, it is a square.

### Acknowledgment

The problem from his book Math Accent has been posted at the CutTheKnotMath facebook page by Dan Sitaru, along with practically identical proofs by Leo Giugiuc, Adil Abdullayev, and Ravi Prakash.