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enter image description here${x^2+y^2+z^2-1}$ this part remembers me to the equation of a sphere, and I can rewrite it as $x^2+y^2+z^2=1$ so the radius of the sphere will be 1. And of coure because of the square root $$\frac{x+y+z-1}{x^2+y^2+z^2-1} \ge0$$

How can I continue it? Do I have to examine where is the numerator and denominator positive/negative?

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Here and as follows, I tried to figure one part of the region, i.e. $x+y+z\geq 1,~~x^2+y^2+z^2>1$. Sice we have serious difficulty to show the later region, I could to regard the box as you see. I am sure you can extend the right region by your self. In

enter image description here

We can see the part of space in which $x+y+z\geq1$. Of coure you may and should extend the red region properly to have it. In

enter image description here

We have the right region in which $x^2+y^2+z^2>1$ and you know why it must not be equal to $1$. Now intersect these two region simultaniously to get one of the region which Ross pointed. I made them by Maple18.

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Yes, the fraction will be positive where the numerator and denominator are both positive or both negative. As you say, $x+y+z-1=0$ is the equation of a plane. The $x+y+z-1$ will be positive on one side of the plane, negative on the other. The denominator represents the unit sphere. It is positive outside the sphere. These divide space into four regions. You need to figure out which two constitute the domain.

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  • $\begingroup$ I still don't know how can I calculate which are these four regions, so this plane "cuts" this sphere and what can I make next? $\endgroup$ – Herrpeter Apr 22 '17 at 14:37
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    $\begingroup$ Your picture is excellent. One region is nearest to us, above the plane and outside the sphere. One is above the plane and inside the sphere. You can just test a point to check the signs in the regions. The origin, for example, is inside the sphere and under the plane. It makes both expressions negative, so is part of the domain. $\endgroup$ – Ross Millikan Apr 22 '17 at 14:51
  • $\begingroup$ I'm still struggling by finding ideal test points. $\endgroup$ – Herrpeter Apr 22 '17 at 14:54
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    $\begingroup$ For the plane, you can just let $x,y,z$ be large and positive, then large and negative. For the sphere, the origin and some place far away are good points. I did the origin above, so the denominator is negative within the sphere and positive outside. $\endgroup$ – Ross Millikan Apr 22 '17 at 14:59

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