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, $P/Q = -\dfrac{9\cdot 4}{2\cdot 81} = -\dfrac{2}{9}.$
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 $$\begin{align}\frac{1}{zx+y-1}+\frac{1}{yz+x-1}+\frac{1}{xy+z-1}&=\frac{y}{y^2-y+4}+\frac{x}{x^2-x+4}+\frac{z}{z^2-z+4}\\\\&=\frac{2}{9}(y-1)+\frac{2}{9}(x-1)+\frac{2}{9}(z-1)\\\\&=\frac{2}{9}(x+y+z-3)\\\\&=\color{red}{-\frac{2}{9}}\end{align}$$
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 $\,S=\cfrac{1}{\frac{9}{2}} \cdot (-1) = - \cfrac{2}{9}\,$.
The following answer avoids any manual computation and reduces the problem to a computation of groebner bases with a computer algebra system. So it is perhaps not quite what was asked for, but it is a good proof of what is possible with Macaulay 2 (or similar systems).
The polynomials $f_1 = x + y + z -2$, $f_2 = x^2 +y^2+z^2 -3$, $f_3 = x y z - 4$ define a zero-dimensional ideal $I$ of degree $6$ in $R = \mathbb{Q}[x,y,z]$, which is invariant under $S_3$. A calculation with Macaulay2 proves that $I$ is prime and so one can define the rings $S = R/I$, an integral domain, and $T=Q(S)$, the quotient field. So every quotient
$$\frac{g(x,y,z)}{h(x,y,z)} \in T$$
invariant under $S_3$ can be reduced to a rational number. The following Macaulay 2 session shows the calculation (with a bonus of getting the value $1/(x^3+y^3+z^3)$ too):
Macaulay2, version 1.9.2
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition,
ReesAlgebra, TangentCone
i1 : R=QQ[x,y,z]
o1 = R
o1 : PolynomialRing
i2 : f1=x+y+z-2
o2 = x + y + z - 2
o2 : R
i3 : f2=x^2+y^2+z^2-3
2 2 2
o3 = x + y + z - 3
o3 : R
i4 : f3=x*y*z-4
o4 = x*y*z - 4
o4 : R
i6 : S=R/ideal(f1,f2,f3)
o6 = S
o6 : QuotientRing
i11 : idI1 = ideal(f1,f2,f3)
2 2 2
o11 = ideal (x + y + z - 2, x + y + z - 3, x*y*z - 4)
o11 : Ideal of R
i12 : primaryDecomposition idI1
2 2 3 2
o12 = {ideal (x + y + z - 2, 2y + 2y*z + 2z - 4y - 4z + 1, 2z - 4z + z - 8)}
o12 : List
i13 : isPrime idI1
o13 = true
i14 : T= frac S
o14 = T
o14 : FractionField
i15 : 1/(z*x+y-1)
-1
o15 = ---------------------
2
y*z + z - y - 2z + 1
o15 : T
i16 : 1/(z*x+y-1)+1/(z*y+x-1)+1/(x*y+z-1)
-2
o16 = --
9
o16 : T
i17 : use R
o17 = R
o17 : PolynomialRing
i18 : dim S
o18 = 0
i19 : degree idI1
o19 = 6
i20 : use T
o20 = T
o20 : FractionField
i21 : 1/(x^3+y^3+z^3)
1
o21 = --
17
o21 : T
