# An Inequality in Cyclic Quadrilateral II

### Source

### Problem

### 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.

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