Irska

Let $F$ stand for a value of the function $F$. Clearly, $x,y,z\in A$ iff the cubic $t^3-3t^2+Ft-1$ has three real roots. Normalise this to the form ($s=t-1$) $$s^3-(3-F)s-(3-F)=0.$$ If $a,b,c$ are the roots of this, then they are real iff $$0\le 4(3-F{)}^3-27(3-F{)}^2=(F-3{)}^2(4(3-F)-27)=-(F-3{)}^2(4F+15).$$ Consequently, the roots are real iff $F=3$ or $F\le -15/4$. Now $F=3$ means that $x,y,z$ satisfy the cubic equation $(t-1{)}^3=0$, and so $x=y=z=1$.

Unless this occurs, then $F\le -\frac{15}{4}$. If there is equality here, then two of $a,b,c$ are equal, i.e., two of $x,y,z$ are equal. Hence, $x=y=-1/2$, $z=4$ say. Thus $$F(A)=\left(-\infty ,-\frac{15}{4}\right]\cup \{3\}.$$