# An Inequality in Cyclic Quadrilateral

### Proof

Rewrite the required inequality as $R(a+b+c+d)^2\ge8\sqrt{2}\sqrt[4]{(abcd)^3}.$

By the AM-GM inequality, $a+b+c+d\ge 16\sqrt[4]{abcd}.\;$ Thus, suffice it to prove that $R\sqrt{2}\sqrt{abcd}\ge\sqrt[4]{(abcd)^3}.\;$ The latter is simplified to

$abcd\le 4R^4.$

Using the AM-GM inequality the second time $\displaystyle abcd\le\frac{(ac+bd)^2}{4}.\;$ By Ptolemy's theorem,

$ac+bd=AC\times BD\le 4R^2,$

so that

$\displaystyle abcd\le\frac{(ac+bd)^4}{4}\le \frac{(4R^2)^2}{4}=4R^4,$

as required. Equality holds only when $a=b=c=d,\;$ i.e., when $ABCD\;$ is a square.

### Acknowledgment

The problem, due to Adil Abdullayev, has appeared at the mathematical inequalities facebook group and was communicated to me, with the above solution, by Leo Giugiuc.