### Solution

$\displaystyle \sqrt{\frac{a^2+b^2+c^2}{3}}\ge\frac{a+b+c}{3},$

implying $\displaystyle \sqrt{a^2+b^2+c^2}\ge\frac{a+b+c}{\sqrt{3}}\;$ and similarly, for other triples of the sides, such that, on adding up, we obtain

$\displaystyle \sum_{cycl}\sqrt{a^2+b^2+c^2}\ge\frac{3(a+b+c+d)}{\sqrt{3}}=\sqrt{3}(a+b+c+d).$

By the AM-GM inequality, $a+c\ge 2\sqrt{ac}\;$ and $b+d\ge 2\sqrt{bd}.\;$ Now, by the Ptolemy's inequality,

$\left(\sqrt{ac}+\sqrt{bd}\right)^2\gt ac+bd\ge AC\cdot BD,$

so that

$\sqrt{ac}+\sqrt{bd}\gt \sqrt{AC\cdot BD}.\;$ Putting everything together shows that

\displaystyle\begin{align} \sum_{cycl}\sqrt{a^2+b^2+c^2} &\ge \sqrt{3}(a+b+c+d)\\ &\ge 2\sqrt{3}(\sqrt{ac}+\sqrt{bd})\\ &\gt 2\sqrt{3}\sqrt{AC\cdot BD}. \end{align}

\displaystyle\begin{align} \end{align}

### Acknowledgment

The problem, due to Dan Sitaru, has been published in the Romanian Mathematical Magazine where more solutions can be found.