Can we assume that $x=y=z$ ?

We simply use a direct application of Holder's Inequality (Which made me wonder, why the high level?)

$\large ((1+5z)(4z+3x)(5x+6y)(y+18))^\frac{1}{4} \geq ((1)(4z)(5x)(y))^\frac{1}{4}+((5z)(3x)(6y)(18))^\frac{1}{4}$ (If you're wondering how the terms got multiplied, try pairing up the terms in such a way that each term gets $xyz$ as a factor)

This simplifies to $\large ((1+5z)(4z+3x)(5x+6y)(y+18))^\frac{1}{4} \geq 4(20xyz)^\frac{1}{4}$

Raising both sides to the $4$ th power gives us $(1+5z)(4z+3x)(5x+6y)(y+18) \geq 5120xyz$

Obtaining the reciprocal of both sides then multiplying both sides by $xyz$ gives us

$\large \frac{zxy}{(1+5z)(4z+3x)(5x+6y)(y+18)} \leq \frac{1}{5120}$

So, our answer is $\boxed{5120}$ .

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P.S. You can check whether equality can hold by the constraint that the ratios of the terms in the parentheses must be constant. Doing so yields $x = \frac{12}{5}$ , $y = 6$ and $z = \frac{3}{5}$ , which is our requirement for the minimum value.

Also, even if the variables can be equal to zero, and Holder's inequality only accepts positive reals, we cannot have a variable equal to zero, since the whole expression will be equal to zero, which is absolutely not a possible maximum.

We begin by establishing a series of inequalities:

$\begin{cases} 1+5z =1+\dfrac{5z}{3}+\dfrac{5z}{3}+\dfrac{5z}{3} \geq 4 \sqrt[4]{\dfrac{125z^3}{27}} \\ 4z+3x =4z+x+x+x \geq 4 \sqrt[4]{4 x^3 z} \\ 5x+6y =5x+2y+2y+2y \geq 4 \sqrt[4]{40 x y^3} \\ y+18 = y+6+6+6 \geq 4 \sqrt[4]{216y} \end{cases}$

Using these inequality on the original inequality gives:

$\dfrac{xyz}{(1+5z)(4z+3x)(5x+6y)(y+18)} \geq \dfrac{xyz}{256 \sqrt[4]{160000x^4 y^4 z^4}}=\dfrac{xyz}{5120xyz}=\dfrac{1}{5120}$

Now this minimum value occurs when $x=\dfrac{12}{5},y=6,z=\dfrac{3}{5}$ so we have:

$\dfrac{1}{r}=\dfrac{1}{5120} \implies \boxed{\boxed{r=5120}}$