For what values of $(x, y)$ does the following inequality hold:

$$
(x+1-y)(x+1+y)(x-1+y)(x-1-y)>0
$$
Multiply the 2nd and 3rd, as well as the 1st and 4th factors:

$$
\left[(x+y)^{2}-1\right]\left[(x-y)^{2}-1\right]>0
$$

The two-factor product on the left side is positive if the signs of the factors are the same. Let's first examine the sign of the first factor on the coordinate plane. If $|x+y|=1$, the value of the expression is 0. The equations $x+y=1$ and $x+y=-1$ are satisfied by the points on two parallel lines. The first line is above the second. Above the first line, $x+y>1$, below the second line, $x+y<-1$, and between the two lines, $-1<x+y<1$. Accordingly, the first factor on the left side of (2) is negative in the strip between the two lines and positive outside the strip.
![](https://cdn.mathpix.com/cropped/2024_05_02_70f09ceec7063e8bc01eg-1.jpg?height=484&width=1108&top_left_y=541&top_left_x=497)

Similarly, we find that the second factor on the left side of (2) is negative in the strip between the parallel lines defined by $x-y=1$ and $x-y=-1$, and positive outside the strip. By superimposing the two strips in our imagination, we find that in the square defined by the points $(1 ; 0)$, $(0 ; 1)$, $(-1 ; 0)$, and $(0 ;-1)$, both factors on the left side of (2) are negative, and in the four corner regions, both are positive. These are the parts of the plane where (1) is satisfied.

![](https://cdn.mathpix.com/cropped/2024_05_02_70f09ceec7063e8bc01eg-1.jpg?height=491&width=526&top_left_y=1294&top_left_x=797)