Evaluate $\frac{1}{zx+y-1}+\frac{1}{zy+x-1}+\frac{1}{xy+z-1}$ if $x+y+z=2,x^2+y^2+z^2=3,xyz=4$ Evaluate $\frac{1}{zx+y-1}+\frac{1}{zy+x-1}+\frac{1}{xy+z-1}$ if $x+y+z=2,x^2+y^2+z^2=3, xyz=4$
The first thing that I notice is that it is symetric to $a,b,c$ but it can't help me .The other idea is finding the numbers but giving it to wolfram alpha gives five complex set of answers. Another idea that looks to be nice is this:
$\sum\limits_{}^{cyc}\frac{1}{xy+z-1}=\sum\limits_{}^{cyc}\frac{x}{x^2-x+4}$
Maybe it gives the answer but it is to hard to calculate. Any hints?

Replacing $x=2-y-z$ and similar, the sum becomes:
$$
S = \sum_{cyc}\frac{1}{x+yz-1} = \sum_{cyc}\frac{1}{1-y-z+yz} =  \sum_{cyc}\frac{1}{(y-1)(z-1)}=\cfrac{1}{\prod_{cyc} (x-1)} \sum_{cyc} (x-1)
$$
From the given relations $\sum_{cyc} xy = \frac{1}{2}\left((\sum_{cyc} x)^2-\sum_{cyc} x^2\right)=\cfrac{1}{2}\,$, so:
$$
\begin{align}
\sum_{cyc}(x-1) & = \sum_{cyc} x - 3 = 2 - 3 = -1 \\
\prod_{cyc}(x-1) & = xyz - \sum_{cyc} xy + \sum_{cyc} x - 1 = 4 - \cfrac{1}{2} + 2 - 1 = \cfrac{9}{2}
\end{align}
$$
Therefore $\,S=\cfrac{1}{\frac{9}{2}} \cdot (-1) = - \cfrac{2}{9}\,$.

P.S. For an alternative derivation, note that $x,y,z$ are the roots of $\,2t^3-4t^2+t-8=0$ by Vieta's formulas. The polynomial with roots $x-1,y-1,z-1$ can be obtained with the substitution $t=u+1$ which, after expanding the powers and collecting, results in $2u^3+2u^2-u-9=0\,$. Therefore $\,\sum_{cyc} (x-1) = -1\,$ and $\,\prod_{cyc} (x-1) = \cfrac{9}{2}\,$.
