<s>[INST] 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?
 [/INST] Edit : Since the OP changes some signs, this answer also changes the signs.
You have already noticed that 
$$\frac{1}{zx+y-1}+\frac{1}{yz+x-1}+\frac{1}{xy+z-1}=\frac{y}{y^2-y+4}+\frac{x}{x^2-x+4}+\frac{z}{z^2-z+4}$$
This answer shows that using that $x,y,z$ are the solutions of $t^3-2t^2+\frac 12t-4=0$ enables us to have a simpler form and to find the value easily.
We have that
$$x+y+z=2,\quad xyz=4$$
and that
$$xy+yz+zx=\frac 12\left((x+y+z)^2-(x^2+y^2+z^2)\right)=\frac{2^2-3}{2}=\frac 12$$
So, we know that $x,y,z$ are the solutions of $$t^3-2t^2+\frac 12t-4=0,$$
i.e.
$$(t-1)(t^2-t+4)-\frac{9}{2}t=0,$$
i.e.
$$\frac{t}{t^2-t+4}=\frac{2}{9}(t-1)$$
since $t^2-t+4\not=0$.
Hence, we get
$$\begin{align}\frac{1}{zx+y-1}+\frac{1}{yz+x-1}+\frac{1}{xy+z-1}&=\frac{y}{y^2-y+4}+\frac{x}{x^2-x+4}+\frac{z}{z^2-z+4}\\\\&=\frac{2}{9}(y-1)+\frac{2}{9}(x-1)+\frac{2}{9}(z-1)\\\\&=\frac{2}{9}(x+y+z-3)\\\\&=\color{red}{-\frac{2}{9}}\end{align}$$
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