FINAL ROUND

It is evident that $$|(a-b)(b-c)(c-d)(d-a)|\le \frac{abcd}{4}⟺$$ $$\frac{(a-b{)}^2}{ab}\cdot \frac{(b-c{)}^2}{bc}\cdot \frac{(c-d{)}^2}{cd}\cdot \frac{(d-a{)}^2}{da}\le \frac{1}{16}.(\ast )$$

Note that $$\frac{(a-b{)}^2}{ab}\le \frac{1}{2}.(1)$$ Indeed, we have $$\begin{gathered} \frac{(a-b{)}^2}{ab}\le \frac{1}{2}\iff 2(a-b{)}^2\le ab\iff 2{\left(\frac{a}{b}\right)}^2-5\left(\frac{a}{b}\right)+2\le 0\iff \\ 2\left(\frac{a}{b}-2\right)\left(\frac{a}{b}-\frac{1}{2}\right)\le 0. \end{gathered}$$ The last inequality holds since $a,b\in [1,2]$. Similarly we have $$\frac{(b-c{)}^2}{bc}\le \frac{1}{2},(2)$$ $$\frac{(c-d{)}^2}{cd}\le \frac{1}{2},(3)$$ $$\frac{(d-a{)}^2}{da}\le \frac{1}{2}.(4)$$ Multiplying (1), (2), (3), (4), we obtain the required inequality (*).

Note that the equality occurs when $$\{(a,b,c,d)=(2,1,2,1),(1,2,1,2)\}.$$