Solve a system of three inequalities in three unknowns I have been trying to solve the system of three inequalities. I could not manage to solve it. Can somebody help me?
$$ X \gt 1 \tag{1} $$
$$ Y \lt Z\cdot (1+X) \tag{2} $$
$$ Z \lt \frac{1-Y}{X} \tag{3} $$
 A: Typically the set of solutions to a system of inequalities will be one or more regions in the space corresponding to the unknowns (here three-dimensional space, for three unknowns).
Working with rational expressions can be especially tricky, since our impulse to multiply both sides to clear fractions is apt to affect the direction of an inequality depending on the sign of what we multiply by.
Here we know $X \gt 1$, so in particular we can multiply by $X$ without changing the direction of any inequality (since $X$ is positive).  The system is then equivalent to this:
$$ X \gt 1 \tag{1} $$
$$ Y \lt ZX + Z \tag{2} $$
$$ ZX \lt 1 - Y \tag{3} $$
It is tempting to multiply the first inequality by $Z$, since then we would be able to replace the original system of nonlinear inequalities by one of strictly linear inequalities, the solution of which is pretty well understood (the subject of linear program feasibility, more or less).
But we don't know in advance whether $Z$ is positive, negative, or zero.  We can resort to considering three separate cases.
Case: Z is zero
If $Z=0$, then the second and third inequalities simplify quite a bit:
$$ Y \lt 0 \tag{2'} $$
$$ 0 \lt 1 - Y \tag{3'} $$
Now it $Y \lt 0$ (rewritten second inequality), then automatically $0 \lt 1-Y$ (rewritten third inequality).  So the case $Z=0$ amounts to a solution set:
$$ \{(X,Y,Z) \mid X\gt 1, Y\lt 0, Z = 0 \} \tag{Z=0} $$
Case: Z is positive
If $Z \gt 0$, then we could multiply the first inequality on both sides by $Z$ without changing the direction of the inequality.  We would then have the equivalent system:
$$ ZX \gt Z \tag{1} $$
$$ Y \lt ZX + Z \tag{2} $$
$$ ZX \lt 1 - Y \tag{3} $$
If we substitute $W = ZX$, then the nature of the equivalent system as linear inequalities will be clear:
$$ W \gt Z \gt 0 \tag{1''} $$
$$ Y \lt W + Z \tag{2''} $$
$$ W \lt 1 - Y \tag{3''} $$
The solutions of the first inequality are easy to visualize.  In the $W,Z$-plane it consists of all the points in the (strict) first quadrant ($W,Z \gt 0$) such that $W \gt Z$, so points $(W,Z)$ falling to the right of the diagonal line $W=Z$.
The second and third inequalities then both amount to upper bounds on $Y$.  That is, not only do we have $Y \lt W + Z$, but also (rewriting the third inequality) $Y \lt 1 - W$.  So the solutions will be (in terms of $W,Y,Z$) determined by:
$$ Y \lt \min(W+Z,1-W) $$
Translating this result back into terms of $X = W/Z$, we have the solution set for the case that $Z$ is positive:
$$ \{(X,Y,Z) \mid X \gt 1, Z \gt 0, Y \lt \min(Z(1+X),1-ZX)\; \} \tag{Z>0}$$
More can be done to visualize this region.  When is $Z(1+X) \le 1-ZX$, and when is the reverse true?  For fixed $Z\gt 0$, the quantity $Z(1+X)$ increases as $X$ increases while $1-ZX$ decreases with increasing $X$.  So for sufficiently large $X$, the smaller of these will be $1-ZX$.
If we work out where the two expressions are equal, $Z(1+X) = 1-ZX$, we find the crossover occurs at:
$$ X_* = \frac{1-Z}{2Z} $$
Since $X\gt 1$, the value $X \ge X_*$ always holds when $Z \ge 1/3$ (which implies $X_* \le 1$), and thus $Y \lt 1 -ZX$ is the limiting constraint.
For $0 \lt Z \lt 1/3$, there are two cases.  If $1\lt X \lt X_*$, then $Y \lt Z(1+X)$ is the limiting constraint, and otherwise ($X \ge X_*$) the limiting constraint is $Y \lt 1 - ZX$.
Case: Z is negative
Left as an exercise for the Reader.  Note that the first inequality becomes $ZX < Z$ because the direction of the inequality is reversed.
