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?

If we transform the whole thing into a single fraction, then both numerator and denominator will be symmetric polynomials on three variables.
The good thing about them is that one can then use the Fundamental Theorem of Symmetric Polynomials to write $P(x,y,z)$ in a unique way as a polynomial $P'(e_1(x,y,z),e_2(x,y,z),e_3(x,y,z))$ where
$$e_1(x,y,z) = x + y + z \qquad e_2(x,y,z) = xy + xz + yz \qquad e_3(x,y,z) = xyz.$$
We already know that $e_1(x,y,z) = 2$ and $e_3(x,y,z) = 4$. We can compute $e_2$ since $e_1^2 - (x^2 + y^2 + z^2) = 2e_2$: 
$$e_2 = \dfrac{1}{2}.$$
Now the only thing we have to do is to decompose the numerator and denominator as a polynomial on $e_1,e_2,e_3$ to find the value of the initial problem. This, however, may not be very elegant.
The whole thing gives $P/Q$ where:
$$P = (x^2yz+xy^2z+xyz^2) + (x^2y+x^2z+xy^2+y^2z+xz^2+yz^2) - (xy+xz+yz) - (2x+2y+2z) + 3,$$
$$Q = (x^3yz+xy^3z+xyz^3) + (x^2y^2z^2) + (x^2y^2+x^2z^2+y^2z^2) -( x^2yz+xy^2z+xyz^2) - (x^2y+x^2z+xy^2+y^2z+xz^2+yz^2) + (x+y+z)  + (xyz)- 1.$$
Therefore
$$x^2yz+xy^2z+xyz^2 = xyz(x+y+z) = 8$$
$$x^2y+x^2z+xy^2+y^2z+xz^2+yz^2 = -3xyz + (xy+xz+yz)(x+y+z) = -11$$
$$2x+2y+2z = 2(x+y+z) = 4$$
So $P = 8 - 11 - \dfrac{1}{2} - 4 + 3 = -\dfrac{9}{2}$.
Conversely,
$$x^3yz + xy^3z + xyz^3 = (xyz)(x^2 + y^2 + z^2) = 12$$
$$x^2y^2z^2 = (xyz)^2 = 16$$
$$x^2y^2+x^2z^2+y^2z^2 - x^2yz-xy^2z-xyz^2 = (xy+xz+yz)^2 - 3(xyz)(x+y+z) = -\dfrac{95}{4}$$
$$x^2y+x^2z+xy^2+y^2z+xz^2+yz^2 = -3xyz + (xy+xz+yz)(x+y+z) = -11$$
so $Q = 12 + 16 -\dfrac{95}{4} + 11 + 2 + 4 - 1  = \dfrac{81}{4}$.
Finally, $P/Q = -\dfrac{9\cdot 4}{2\cdot 81} = -\dfrac{2}{9}.$
