Show that the integral can not exceed $\frac{\pi^2}{96}$ Show that
$$
\int_{-\infty}^{\infty}\int_{-x}^{x}\int_{-y}^{y}\int_{-z}^{z}e^{-(x^2+y^2+z^2+w^2)}\dfrac{|zw|}{(1+x^2)(1+y^2)} \,dw \,dz\,dy\,dx\le\frac{\pi^2}{96}
$$
I am not understanding how $\pi$ is coming into picture, and I'm not sure I can get the upper bound.
 A: By symmetry, then killing a couple of variables:
$$\begin{eqnarray*} I &=& 16\int_{0}^{+\infty}\int_{0}^{x}\int_{0}^{y}\int_{0}^{z}\frac{zw e^{-(z^2+w^2)}}{(1+x^2)(1+y^2)}e^{-(x^2+y^2)}\,dw\,dz\,dy\,dx\\&=&8\int_{0}^{+\infty}\int_{0}^{x}\int_{0}^{y}\frac{z e^{-z^2}(1-e^{-z^2})}{(1+x^2)(1+y^2)}e^{-(x^2+y^2)}\,dz\,dy\,dx\\&=&2\int_{0}^{+\infty}\int_{0}^{x}\frac{(1-e^{-y^2})^2}{(1+x^2)(1+y^2)}e^{-(x^2+y^2)}\,dy\,dx\end{eqnarray*} $$
so by upper-bounding $(1-e^{-y^2})^2$ with $1$ and lower-bounding $e^{x^2}$ with $(1+x^2)$ we get:
$$ I\leq \int_{0}^{+\infty}\int_{0}^{+\infty}\frac{e^{-x^2}\cdot e^{-y^2}}{(1+x^2)(1+y^2)}\,dx\,dy = \left(\int_{0}^{+\infty}\frac{dx}{e^{x^2}(1+x^2)}\right)^2 \leq \left(\int_{0}^{+\infty}\frac{dx}{(1+x^2)^2}\right)^2\leq \frac{\pi^2}{16}.$$
The missing $\frac{1}{6}$ factor probably comes from a more cunning management of the $(1-e^{-y^2})^2$ term.
Using the fact that $(1-e^{-y^2})^2 e^{-y^2}\leq\frac{4}{27}$ (that follows from the AM-GM inequality) we have:
$$ I\leq \frac{8}{27}\int_{0}^{+\infty}\int_{0}^{x}\frac{1}{(1+y^2)(1+x^2)^2}\,dy\,dx = \frac{\pi^2-4}{54}$$
that is way closer to the inequality we want to prove. Using $(1-e^{-y^2})^2 e^{-y^2}\leq\min\left(y^4,\frac{4}{27}\right)$ we should improve the original inequality, so $\frac{\pi^2}{96}$ probably just comes from a probabilistic/Parseval/Cauchy-Schwarz argument buried in the thoughts of the problem poser.
