Calculate the sum of fractions

[anemone]

If \(\displaystyle x,\;y,\;z\in\mathbb{C}\) and \(\displaystyle x+y+z=2\), \(\displaystyle x^2+y^2+z^2=3\), \(\displaystyle xyz=4\), calculate \(\displaystyle \frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}\).

 

[Opalg]

If \(\displaystyle x,\;y,\;z\in\mathbb{C}\) and \(\displaystyle x+y+z=2\), \(\displaystyle x^2+y^2+z^2=3\), \(\displaystyle xyz=4\), calculate \(\displaystyle \frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}\).

Spoiler

Write $\sum x$ for $x+y+z$, $\sum yz$ for $yz+zx+xy$ and $\sum x^2$ for $x^2+y^2+z^2$. Then $2\sum yz = \Bigl(\sum x\Bigr)^2 - \sum x^2 = 2^2-3 = 1$ and so $\sum yz = \frac12.$

The equation with roots $x,y,z$ is thus $t^3 - 2t^2 + \frac12t - 4 = 0$. (This equation only has one real root, so it's just as well that the question is about complex numbers.)

Now let $\alpha = yz+x$, $\beta = zx+y$ and $\gamma = xy+z$. The next step is to find the symmetric functions $\sum\alpha$, $\sum\beta\gamma$ and $\alpha\beta\gamma$. For this, we first need to find the value of $\sum x\bigl(y^2+z^2\bigr)$ and $\sum y^2z^2$. To do that, notice that $\sum x \sum yz = 3xyz + \sum x\bigl(y^2+z^2\bigr)$, from which $\sum x\bigl(y^2+z^2\bigr) = -11.$ Also, $\Bigl(\sum yz\Bigr)^2 = \sum y^2z^2 + 2xyz\sum x$, from which $\sum y^2z^2 = \frac14 - 16 = -\frac{63}4.$

We can now calculate that $$\textstyle \sum\alpha = \sum yz + \sum x = \tfrac72,$$ $$\textstyle \sum\beta\gamma = \sum\bigl(x^2yz + xy^2 + xz^2 + yz\bigr) = xyz\sum x + \sum x\bigl(y^2+z^2\bigr) + \sum yz = 8 - 11 + \tfrac12 = -\tfrac52,$$ $$\textstyle \alpha\beta\gamma = (yz+x)(zx+y)(xy+z) = (xyz)^2 + xyz\sum x^2 + \sum y^2z^2 + xyz = 16 + 12 - \tfrac{63}4 + 4 = \tfrac{65}4.$$

From those calculations, the equation with roots $\alpha,\,\beta,\,\gamma$ is $t^3 - \frac72t^2 - \frac52t - \frac{65}4$. The equation with roots $\alpha-1,\,\beta-1,\,\gamma-1$ is $(t+1)^3 - \frac72(t+1)^2 - \frac52(t+1) - \frac{65}4$, or $4t^3 -2t^2 -26t - 85 = 0.$ Replacing $t$ by $1/t$, it follows that the equation with roots $1/(\alpha-1),\,1/(\beta-1),\,1/(\gamma-1)$ is $85t^3 + 26t^2 + 2t - 4 = 0.$

Finally, the sum of the roots of that last equation is $$-\frac{26}{85} = \frac1{\alpha-1} + \frac1{\beta-1} + \frac1{\gamma-1} = \frac1{yz+x-1} +\frac1{zx+y-1} + \frac1{xy+z-1}.$$

Disclaimer: I have not checked the details, so the answer may be wrong. But the method should be correct.

Edit. That answer is indeed wrong, and the answer given below by Jester and anemone is correct. I don't have the patience to search for where I went wrong, and in any case anemone's method is far neater and shorter than mine.

 

[Mathwebster]

I did it another way (and got a different answer).

Spoiler

First, as was stated $x, y$ and $z$ satisfy

$2r^3 - 4r^2 + r - 8 = 0$

Next, we can re-write our target equation using $xyz=4$ as

$ \dfrac{1}{\dfrac{4}{x} + x - 1} + \dfrac{1}{\dfrac{4}{y} + y - 1} + \dfrac{1}{\dfrac{4}{z} + z - 1}$

If we let $w = \dfrac{1}{\dfrac{4}{x} + x - 1}$ then we can show that

$ w^3 + \dfrac{2}{9}w^2 - \dfrac{2}{81}w - \dfrac{4}{81} = 0$

so the sum of the three roots, which is our target, is $-\dfrac{2}{9}$.

 

[anemone]

Thanks to Opalg and Jester for participating in this challenge problem.

I solved it differently and here is my solution:

Spoiler

\(\displaystyle \frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}\)

\(\displaystyle =\frac{1}{xy+(2-x-y)-1}+\frac{1}{yz+(2-y-z)-1}+\frac{1}{xz+(2-x-z)-1}\)

\(\displaystyle =\frac{1}{xy-x-y+1}+\frac{1}{yz-y-z+1}+\frac{1}{xz-x-z+1}\)

\(\displaystyle =\frac{1}{(x-1)(y-1)}+\frac{1}{(y-1)(z-1)}+\frac{1}{(x-1)(z-1)}\)

\(\displaystyle =\frac{(z-1)+(x-1)+(y-1)}{(x-1)(y-1)(z-1)}\)

\(\displaystyle =\frac{2-3}{xyz-(xy+yz+xz)+(x+y+z)-1}\)

\(\displaystyle =\frac{-1}{4-\frac{1}{2}+(2)-1}\)

\(\displaystyle =-\frac{2}{9}\)

 

[anemone]

Hi Jester,

Spoiler

I have tried many ways to figure out how could we end up with the cubic function of $ w^3 + \dfrac{2}{9}w^2 - \dfrac{2}{81}w - \dfrac{4}{81} = 0$ if we let $w = \dfrac{1}{\dfrac{4}{x} + x - 1}$...with no luck.

My convoluted attempt to generalize that cubic function is a great deal more difficult than to find the value for the target expression, LOL!

Could you please tell me how did you obtain $ w^3 + \dfrac{2}{9}w^2 - \dfrac{2}{81}w - \dfrac{4}{81} = 0$ as you showed in your solution?

Many, many thanks in advance.

 

[Mathwebster]

Spoiler

With my target expression, since each expression is the same, I figured that if I could find a cubic associated with it, I would have the answer I was looking for. Given that x satisfies

$2x^3-4x^2+x-8 = 0$

then it was a matter of tour de force

Letting $w =\dfrac{1}{\dfrac{4}{x} + x-1}$ and allowing that $w$ satisfy a cubic

$w^3 + aw^2 + bw + c = 0$

Simplifying gave

$c{x}^{6}+ \left( b-3\,c \right) {x}^{5}+ \left( -2\,b+a+15\,c \right)
{x}^{4}+ \left( 1-a+9\,b-25\,c \right) {x}^{3}$
$+ \left( -8\,b+4\,a+60\,
c \right) {x}^{2}+ \left( 16\,b-48\,c \right) x+64\,c=0$.

Imposing the cubic in $x$ reduced this to

$\left( 13\,b+2+\frac{11}{2}\,a+{\frac {235}{4}}\,c \right) {x}^{2}+ \left( {\frac {47}{4}}\,b-\frac{1}{2}-\frac{1}{4}\,c+\frac{7}{2}\,a \right) x+4+34\,b+82\,c+4\,a = 0
$

and setting the coefficients to zero and solving for $a, b$ and $c$ gave me what I was looking for.

 

[anemone]

Hi Jester,

Great explanation and thanks for the insights!:)