Evaluate The Sum Of 1/xy+z-1+1/yz+x-1+1/xz+y-1

The first problem I am going to post here is perfect for training/study and to stretch and inspire the most able and willing secondary school pupils and help to develop one's ability and harness enthusiasm for mathematics.

Given [MATH]x,\;y,\;z\in\mathbb{C}[/MATH]

where:

[MATH]\begin{cases}x+y+z=2 \\[3pt] x^2+y^2+z^2=3 \\[3pt] xyz=4 \\ \end{cases}[/MATH]

Calculate [MATH]\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}[/MATH].

If one decides to blindly push ahead to solve this challenging math problem, one may be tempted proceed by first adding up the three fractions and then continue from there:

[MATH]\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}[/MATH]

[MATH]=\frac{(yz+x-1)(xz+y-1)+(xy+z-1)(xz+y-1)+(xy+z-1)(yz+x-1)}{(xy+z-1)(yz+x-1)(xz+y-1)}[/MATH]

Okay, if you continue working the problem from here, you would wind up with:

[MATH]=\dfrac{(xyz^2+y^2z+-yz+x^2z+xy-x-xz-y+1)+(x^2yz+xy^2-xy+xz^2+yz-z-xz-y+1)+(xy^2z+x^2y-xy+yz^2+xz-z-yz-x+1)}{(xy+z-1)(yz+x-1)(xz+y-1)}[/MATH]

Now, you might realize something seems wrong because it appears to be hard to decide which terms to group, but you are a tenacious bull, and so you want to continue, and you try this:

[MATH]=\dfrac{(xyz^2+y^2z++x^2z+xy-(xz+yz)+1-(x+y))+(x^2yz+xy^2+xz^2+yz-(xy+xz)+1-(y+z))+(xy^2z+x^2y+yz^2+xz-(xy+yz)-(x+z)+1)}{(xy+z-1)(yz+x-1)(xz+y-1)}[/MATH]

I think you will agree with me that you would stop working at this point in time as you feel you have wound up in a blind alley.

You know by now that one could never approach this problem using an "ordinary method", indeed, one needs to think first to see if we could, as we go, simplify the expressions, and so we will start to work the problem from the beginning:

We need to calculate:

[MATH]\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}[/MATH].

where:

[MATH]\begin{cases}x+y+z=2 \\[3pt] x^2+y^2+z^2=3 \\[3pt] xyz=4 \\ \end{cases}[/MATH]

Okay, we have to use the given information wisely so it serves to assist us:

Note that we are given $x+y+z=2$, we can rewrite this to make the $z$ the subject, and get $z=2-(x+y)$ and by the same token, $x=2-(y+z)$ and $y=2-(x+z)$.

Substitute into the given expression and recognize that $xy-x-y+1=(x-1)(y-1)$, so that we get:

[MATH]\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}[/MATH]

[MATH]=\frac{1}{xy+(2-x-y)-1}+\frac{1}{yz+(2-y-z)-1}+\frac{1}{xz+(2-x-z)-1}[/MATH]

[MATH]=\frac{1}{xy-x-y+1}+\frac{1}{yz-y-z+1}+\frac{1}{xz-x-z+1}[/MATH]

[MATH]=\frac{1}{(x-1)(y-1)}+\frac{1}{(y-1)(z-1)}+\frac{1}{(x-1)(z-1)}[/MATH]

[MATH]=\frac{(z-1)+(x-1)+(y-1)}{(x-1)(y-1)(z-1)}[/MATH]

[MATH]=\frac{2-3}{xyz-(xy+yz+xz)+(x+y+z)-1}[/MATH]

[MATH]=\frac{-1}{4-\frac{1}{2}+(2)-1}[/MATH]

[MATH]=-\frac{2}{9}[/MATH]

Wow..this has been solved so beautifully! The only "magic" that helped us to solve this problem neatly and elegantly was to recognize and bear in mind that $xy-x-y+1=(x-1)(y-1)$ and everything else fell into place as it should. Amazing, isn't it?

Have you begun to love challenging math problems? Please don't forget that in order to appreciate a language, you must become fluent.