What could be the solution of the inequality $(x - y)(x + y - 1) > z$?

Question

I need to find the solution to the inequality $(x - y)(x + y -1) > z$, where $x,y,z \geq 0$ and $x,y,z \leq 1$. As $z$ is positive, then the inequality holds whenever (i) $x - y > 0$ and $x + y - 1 > 0$ OR (ii) $x - y < 0$ and $x + y - 1 < 0$. I can solve the cases (i) and (ii) on their own, but I don't know how to relate them to the value of $z$. Probably the solution is simple, but I am not an expert in math so any help would be appreciated.

Answer 1

If we multiply and rearrange the terms in the inequality, we get: $$z < x^2 - y^2 - x + y$$ which is equivalent to: $$z < \left (x - \frac 1 2 \right)^2 - \left (y - \frac 1 2 \right)^2.$$ The associated equation represents a hyperbolic paraboloid having its saddle point at $\left ( \frac 1 2 , \frac 1 2, 0 \right )$.

If we rewrite the inequality as the system: $$\begin{cases} \left (x - \frac 1 2 \right)^2 - \left (y - \frac 1 2 \right)^2 > k \\ z = k \end{cases} \qquad (k \in \mathbb R)$$ we see that the inequality represents the set $S$ of all points that, at a given altitude $k$, lie outside the corresponding hyperbola.

Now, if we also require that $0 \le x, y, z \le 1$, we are just intersecting $S$ with the cube centered at $\left ( \frac 1 2, \frac 1 2, \frac 1 2 \right)$ having side of length $1$. This means that we consider only the points having altitude $k$ with $0 \le k \le 1$ that lie on the inside of the square having vertices $(0, 0, k)$, $(1, 0, k)$, $(1, 1, k)$, $(0, 1, k)$ and on the outside of the corresponding hyperbola.

We can look at some of those points for given altitudes.

For $k = 0$ we get:

For $k = \frac 1 {20}$, for instance, we get:

and so on, until for $\frac 1 4 \le k \le 1$ we get:

because for $k = \frac 1 4$ the hyperbola has its vertices at $(0, \frac 1 2, k)$ and $(1, \frac 1 2, k)$.

Answer 2

Hint; It is $$x^2-x-y^2+y-z>0$$ you can solve it now for $x$ or $y$.