一道四元代数不等式

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$a+b+c+d=1,a>0,b>0,c>0,d>0$,求$(b c - a d) (c a - b d) (a b - c d)$的范围.
最小值应该是$-\frac{1}{4096}$, 当$a\to \frac{1}{8},b\to \frac{5}{8},c\to \frac{1}{8},d\to \frac{1}{8}$时能取等
最大值是$\frac{1}{729}$,当$a\to \frac{1}{3},b\to 0,c\to \frac{1}{3},d\to \frac{1}{3}$时能取等.

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Possibly Related: 一道四元对称不等式
$(b c - a d) (c a - b d) (a b - c d) = (a b c + b c d + c d a +  d a b)^2-a b c d (a + b + c + d)^2$