Ireland

By the AM-GM inequality, $$\begin{array}{rl}& abc+bcd+dca+dab\ge 4\sqrt[4]{(abcd{)}^3}\text{and}\\ & ab+bc+cd+da+ac+bd\ge 6\sqrt[6]{(abcd{)}^3},\end{array}$$ with equality in both inequalities iff $a=b=c=d$. Hence, letting $x=\sqrt[4]{abcd}$, we have that $$1\ge 3x^4+8x^3+6x^2$$ with equality iff $a=b=c=d$. But $1-3x^4-8x^3-6x^2=(1-3x)(1+x{)}^3$, and $x$ is positive. Hence, $1\ge 3x^4+8x^3+6x^2$ iff $x\le 1/3$, with equality iff $x=1/3$. Thus $abcd\le 1/3^4$, and the inequality is strict unless $a=b=c=d=1/3$.