Answer
Please see the graph.
Work Step by Step
The green line is the graph $x^{2}-y^{2}\ge1$ and has a dotted line since it has a less than sign. The blue line is the graph $\frac{x^{2}}{16}+\frac{y^{2}}{4}\le1$ and has a solid line since it has a less than or equal to sign. The orange line is the graph $y\ge1$ and has a solid line since it has a greater than or equal to sign.
We also pick the point $(0,2)$ to determine what sides of the graphs to shade. For the green graph, we need three points to determine what sides to shade since it is a hyperbola. We pick $(-2,0)$, $(0,0)$, and $(2,0)$.
$(-2,0)$
$ x^{2}-y^{2}\ge1$
$(-2)^{2}-0^{2}\ge1$
$4-0 \ge 1$
$4 \ge 1$ (true, so we shade the side of the graph with the point)
$(0,0)$
$ x^{2}-y^{2}\ge1$
$0^{2}-0^{2}\ge1$
$0-0 \ge 1$
$0 \ge 1$ (false, so we shade the side of the graph without the point)
$(2,0)$
$ x^{2}-y^{2}\ge1$
$(2)^{2}-0^{2}\ge1$
$4-0 \ge 1$
$4 \ge 1$ (true, so we shade the side of the graph with the point)
$(0,2)$
$\frac{x^{2}}{16}+\frac{y^{2}}{4}\le1$
$\frac{0^{2}}{16}+\frac{2^{2}}{4}\le1$
$0/16 + 4/4 \le 1$
$1 \le 1$ (true, so we shade the side of the graph with the point)
$(0,2)$
$y\ge1$
$2\ge1$
$2 \ge 1$ (true, so we shade the side of the graph with the point)
The overlap of the graphs is the solution set.
