How to evaluate this limit involving a triple integral? How to evaluate this limit involving a triple integral?
\begin{align*}
\lim_{n\to \infty}\sqrt[n]{\int_0^1\int_0^1\int_0^1\frac{x^n(1-x)^ny^n(1-y)^nz^n(1-z)^n}{[1-(1-xy)z]^{n+1}}d x dydz}
\end{align*}
It's easy to see that
\begin{align*}
&\sqrt[n]{\int_0^1\int_0^1\int_0^1\frac{x^n(1-x)^ny^n(1-y)^nz^n(1-z)^n}{[1-(1-xy)z]^{n+1}}d xd yd z}\\
\leqslant &
\sqrt[n]{\int_0^1\int_0^1\int_0^1\frac{x^n(1-x)^ny^n(1-y)^nz^n(1-z)^n}{[1-(1-xy)][1-z]^{n}}d xd yd z}\\
\leqslant &
\sqrt[n]{\int_0^1\int_0^1\int_0^1{x^{n-1}(1-x)^ny^{n-1}(1-y)^nz^n(1-z)^n}d xd yd z}\\
=&\sqrt[n]{\int_0^1x^{n-1}(1-x)^nd x\int_0^1y^{n-1}(1-y)^nd y\int_0^1{z^n}d z}
\to \frac{1}{4}\cdot \frac14\cdot 1=\frac1{16},n\to \infty \\
&~(\text{By using Stirling's formula})
\end{align*}
But I'm not sure $\frac1{16}$ is the right answer.How to do it ?Can anyone afford some help?Thanks in advance.
 A: This is not an answer.
Just out of curiosity, I computed numerically the value of 
$$
f_n=\sqrt[n]{\int_0^1\int_0^1\int_0^1\frac{x^n\,(1-x)^n\,y^n\,(1-y)^n\,z^n\,(1-z)^n}{[1-(1-xy)z]^{n+1}}\,dx\, dy\,dz}
$$ and obtained the following results
$$\left(
\begin{array}{cc}
 n & f_n \\
 5 & 0.0453783835 \\
 10 & 0.0568806730 \\
 15 & 0.0631494275 \\
 20 & 0.0672030029 \\
 25 & 0.0700772984 \\
 30 & 0.0722394764 \\
 35 & 0.0739346466 \\
 40 & 0.0753039687 \\
 45 & 0.0764360143 \\
 50 & 0.0773910546
\end{array}
\right)$$
A: Let $\mathbb D$ be the unit cube in $\mathbb R^3$ (a cube with volume $1$ whose corner is at the origin) and define $g:\mathbb D\to\mathbb R$ as:
$$g(\mathbf x)=\lim_{n\to\infty}\frac{x_1x_2x_3(1-x_1)(1-x_2)(1-x_3)}{(1-x_3+x_1 x_2 x_3)^{1+\frac 1n}}$$
If we define $f:\mathbb R^3\to\mathbb R$ as
$$f(\mathbf x)=\begin{cases}g(\mathbf x)&\mathbf x\in\mathbb D\\0&\text{elsewhere}\end{cases}$$
then your limit is the definition of the infinity-norm of $f$:
$$\|f\|_\infty=\lim_{n\to\infty}\left(\int_V |f|^ndv\right)^{\frac 1n}$$
which, if I am not mistaken, is equal to:
$$\|f\|_\infty=\sup\{|f(x)|:x\in\mathbb{R}^3\}$$
So we are left with calculating the suprimum of $g$ on $\mathbb D$, which occurs at:
$$(x_1,x_2,x_3)=\left(\sqrt{2}-1,\sqrt{2}-1,\frac{\sqrt{2}}{2}\right)\implies g(x)=17-12\sqrt{2}$$
Edit- The above values are obtained by running this code in Mathematica:
Maximize[{x1 x2 x3(1-x1)(1-x2)(1-x3)/(1-x3+x1 x2 x3),0<x1<1 && 0<x2<1 && 0<x3<1},{x1,x2,x3}]

The formal way is to calculate the derivative of $g$ and see where it vanishes. Of course that would be tedious.
