SAUDI ARABIAN MATHEMATICAL COMPETITIONS

First, if $z=\frac{1}{y}$, then the given system becomes $$\{\begin{array}{c}x\cdot y\cdot \frac{1}{y}>1\\ x+y+\frac{1}{y}>\frac{1}{x}+\frac{1}{y}+y\end{array}\iff x>1.$$ So, all triples $\left(x,y,\frac{1}{y}\right)$, with $x>1$ and $y\ne 0$, satisfy the given system of inequalities.

Next, note that the range of $f(y):=y+\frac{1}{y}$ is $(-\infty ,-2]\cup [2,+\infty )$ and that, for such triples, $S=x+f(y)$. Then for each value $a\in \mathbb{R}$, we need only consider the following cases:

1. If $a>-1$, we can choose $x=a+2(>1),y=-1$, and get $S=a+2+f(-1)=a$.

2. If $a<-1$, we let $x=-a(>1)$ and choose $y\ne 0$ such that $f(y)=2a(\in (-\infty ,-2])$. Then $S=-a+f(y)=a$.

3. If $a=-1$, we choose $x=1.5(>1),y=-2$ and get $S=1.5+f(-2)=-1=a$.

Therefore, $S$ can take on any real values when $x,y,z$ vary.