<s>[INST] Domain of $f(x,y) = {\sqrt{x+y-1 \over x-y+1}}$ How I can get the domain of the function
$$ f(x,y) = \sqrt{x+y-1 \over x-y+1}?$$
I know that is: $x+1 \neq y$ and $x^2 \ge (y-1)^2$
But I don't know how to get the second condition.
 [/INST] We need $\dfrac{x+y-1}{x-y+1} \ge 0$.
1) $x+y-1 \ge 0$, and $x-y+1 >0$
$(x+(y-1)) \cdot (x-(y-1)) \ge 0$;
$x^2 \ge (y-1)^2$;
2) $x+y-1\le 0$, and $x -y +1 <0;$
Proceed likewise, note that multiplying 
$(x+y-1) \le 0$ by $(x-y+1) <0$
changes the sign of the inequality.
$(x +(y-1))(x-(y-1)) \ge 0.$
Same result as in 1).
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