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.
 A: Do you know that $\frac{f(x)}{g(x)}\ge0\implies f(x)\cdot g(x)\ge0$ provided that $g(x)\ne0$? So, here's how you get the second condition (the square root function is only valid for real numbers that are greater than or equal to zero):
$$\require{cancel}
\frac{x+y-1}{x-y+1}\ge0\implies\\
(x+y-1)(x-y+1)\ge0\implies\\
[x+(y-1)][x-(y-1)]\ge0\implies\\
x^2-(y-1)^2\ge0\implies\\
x^2\ge(y-1)^2.
$$
A: Hint: Where defined, $f(x,y)$ equals
$$f(x,y)=\sqrt{\frac{x+y-1}{x-y+1}\cdot\frac{x-y+1}{x-y+1}}=\frac{\sqrt{(x+(y-1))\cdot(x-(y-1))}}{|x-y+1|}=\frac{\sqrt{x^2-(y-1)^2}}{|x-y+1|}. $$
A: $f$ is undefined only in 2 cases -


*

*Denominator is $0$

*The term in the square root is negative (which itself is only possible if the numerator and denominator has opposite signs)


Proceed from here.
A: 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).
