Minimize 3 Variable Product

Equality case is

$(x,y,z)=\left(2,1,\dfrac{1}{2}\right)$

Which gives a value of 27.

Hint: Use Reverse Rearrangement Inequality

EDIT: here is a full solution with motivation.

Given the hint of using RR, along with the restriction $0\le x\le 2y\le 4z$ , we consider using the sequence $\{x, 2y, 4z\}$ as one of the sequences of reverse rearrangement. Clearly, since we want to find the minimum value and we are given $xyz=1$ , we should have the other sequence be $\{ax, by, cz\}$ for some real values $a,b,c$ . Since the expression we want to minimize has an $8z$ in it, we guess $c=8$ . Also, to make the constants factor out, we guess $a=2$ and $b=4$ . Note that $2x\le 4y\le 8z$ so we can use RR.

We use RR: $(x+8z)(2y+2x)(4z+4y)\ge (x+2x)(2y+4y)(4z+8z)$ $\iff 8(x+y)(y+z)(x+8z)\ge 3\cdot 6\cdot 12\cdot xyz$ $\iff (x+y)(y+z)(x+8z)\ge 27$

Thus the answer is $\boxed{27}$