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if :

$x+y+z=2$ , $ x^2+y^2+z^2=3$ , $xyz=4$

Then evaluate:

$\frac {1} {xy+z-1} + \frac {1} {yz+x-1} + \frac {1} {zx+y-1}$

My try:

$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=4 \rightarrow 3+2(xy+yz+xz)=4 \rightarrow xy+yz+xz=\frac {1}{2} \rightarrow xy=\frac {1}{2}-yz-xz$

$\rightarrow xy+z-1=\frac {1}{2}-yz-xz+z-1=-\frac{1} {2} +z(1-y-x)$ $\mathbb{I}$

$x+y+z=2 \rightarrow z-1=1-y-x$ $\mathbb{II}$

$\mathbb{I,II} \rightarrow -\frac {1}{2}+z(z-1)=z^2-z-\frac {1}{2}$ $\mathbb{III}$

$xyz=4 \rightarrow xy=\frac {4}{z} \rightarrow xy+z-1=\frac {4}{z} +z -1=\frac {4+z^2-z}{z}$ $\mathbb{IV}$

$\mathbb{III,IV} \rightarrow z^2-z-\frac{1}{2}=\frac {4+z^2-z}{z} \rightarrow \text{after simplifying} \rightarrow z^3-2z^2+\frac{1}{2} z -4=0$

From that I can find the value of $z$ and doing the same calculations again, values of $y,x$ can be found.

I can't seem to be able to proceed from here and a really strong hunch tells me that I've done a terrible mistake somewhere here.Besides I don't really know how the equation above is solved without using a computer.

So please don't post the complete answer here what I'm asking for is:

1-A hint or anything that can lead me to the solution

2-Finding the error in my solution (if there's any)

3-Please don't laugh at me if I've made a really really stupid mistake back there because I've spent a nice 3 hours on this and it really hurts. :D

P.S:I myself couldn't identify the error in my calculation I have checked them more that a few times.Also please feel free to edit the title and the tags , but I'm pretty sure the question itself has been written correctly.(According to the book here I mean)

Thank you all for your time and help in advance.

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  • $\begingroup$ Multiplying and dividing each fraction by z,x and y respectively to get $\sum \frac{z}{z^2 - z + 4}$. Add them up and simplify it using the given 3 equations. $\endgroup$ – Bhaskar Apr 20 '15 at 14:20
  • $\begingroup$ @L16H7 Thanks but I also need to know where i've went wrong in my solution. $\endgroup$ – Arian Tashakkor Apr 20 '15 at 14:42
  • $\begingroup$ What you are doing is you are forming a cubic equation with x,y,z as its roots. If from cubic equation you can find the roots easily, then this method would be fine. I don't think there's any mistake in your calculations. $\endgroup$ – Bhaskar Apr 20 '15 at 14:52
  • $\begingroup$ @L16H7 Thank you.Can you elaborate your hint again please I don't think I've understood what you said $\endgroup$ – Arian Tashakkor Apr 20 '15 at 15:15
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$\bf{My\; Solution::}$ Given $$x+y+z = 2\;\;, x^2+y^2+z^2=3\;\;,xyz = 4$$

Now $$xy+z-1 = xy+2-x-y-1 = (1-x)\cdot (1-y)......................(1)\color{\red}\checkmark$$

Above we put $z=2-x-y$ in $xy+z-1$

Similarly $$yz+x-1=yz+2-y-z-1 = (y-1)\cdot (z-1).........................(2)\color{\red}\checkmark$$

Similarly $$zx+y-1=zx+2-x-z-1 = (z-1)\cdot (x-1).........................(3)\color{\red}\checkmark$$

So $$\displaystyle \frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{zx+y-1} = \frac{1}{(1-x)\cdot (1-y)}+\frac{1}{(y-1)\cdot (z-1)}+\frac{1}{(z-1)\cdot (x-1)}$$

So Expression $$\displaystyle = \frac{3-(x+y+z)}{(1-x)\cdot (1-y)\cdot (1-z)} = \frac{3-(x+y+z)}{1-(x+y+z)+(xy+yz+zx)-xyz}$$

Now Using $$\displaystyle (x+y+z)^2=(x^2+y^2+z^2)+2(xy+yz+zx)\Rightarrow xy+yz+zx = \frac{1}{2}.$$

Put These value in above equation , We Get $$\displaystyle = \frac{3-2}{1-2+\frac{1}{2}-4} = -\frac{2}{9}$$

So $$\displaystyle \frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{zx+y-1} = -\frac{2}{9}$$

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  • $\begingroup$ Much appreciated!Elegant and simple.Although I would've been much more thankful if you'd just give a hint. :D But, many thanks!BTW can you check my try as well see if it also could lead to something? $\endgroup$ – Arian Tashakkor Apr 20 '15 at 17:17

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