I'm trying to compute the sum of this infinite series, but I can't really find a common ratio for it. I've tried breaking it up into the sum of three separate series but to no avail. I'm not sure that I understand exactly how the series works. I need to put it into a form that has $\pi$, but I only lead on that was with the Riemann zeta function. Here's the sum.
$$\sum_{x=1}^{\infty}\sum_{y=1}^{\infty}\sum_{z=1}^{\infty}\frac{100z^2yx}{z^3y^4x^3+z^3y^3x^4+z^4y^3x^3}$$
On
Forget the 100 for a bit.
Clearly you will get the same answer if you permute $x,y,z$. So the sum is one-third of the same sum with $x^2 yz +y^2 zx +z^2 xy$ on top.
Then the term we are summing reduces at once to $\frac{1}{x^2}\frac{1}{y^2}\frac{1}{z^2}$.
So we'll get something like $\frac{100}{3}(\frac{\pi^2}{6})^3$.
This is only a partial answer with which you might want to proceed. Let me denote your sum by $S$. Then \begin{align} S &= \sum_{x=1}^{\infty}\sum_{y=1}^{\infty}\sum_{z=1}^{\infty}\frac 1{zy^2x^2}\cdot\frac{100}{x+y+z}\\ &= 100\sum_{x=1}^{\infty}\frac 1{x^2}\sum_{y=1}^{\infty}\frac 1{y^2}\sum_{z=1}^{\infty}\frac 1{z}\cdot\frac{1}{x+y+z}\\ &= 100\sum_{x=1}^{\infty}\frac 1{x^2}\sum_{y=1}^{\infty}\frac 1{y^2}\frac{1}{x+y}\sum_{z=1}^{\infty}\left(\frac{1}{z}-\frac{1}{x+y+z}\right)\\ &=100\sum_{x=1}^{\infty}\frac 1{x^2}\sum_{y=1}^{\infty}\left(\frac{1}{x+y}\sum_{z=1}^{x+y}\frac{1}{z}\right)\frac 1{y^2}\\ &=100\sum_{x=1}^{\infty}\sum_{y=1}^{\infty}\frac{M(x+y)}{(xy)^2}, \end{align} where $M(n) = \frac{1}{n}\sum_{k=1}^{n}\frac{1}{k}$.
EDIT: I just saw that setting $p=x+y$ you get $$ S = \sum_{p=2}^\infty M(p)\sum_{x=1}^{p-1}\frac 1{x^2(p-x)^2}. $$ Maybe you can do the partial fraction decomposition on the inner series again to simplify it... I have to go now. Good luck!