The hyperbolic set in my problem is defined as:
$H = \{x \ \epsilon \ R_+^2 \ | \ x_1x_2 \ \geq \ 1\}$
I need to show that this can be written as an intersection of halfspaces. In the previous part I showed that it is convex using first principles so there must exist such an intersection of halfspaces that gives us this set but I'm not sure where to start. What exactly constitutes as an intersection of halfspaces?
Let me use $x,y$ coordinates, and so your set is $$H = \{(x,y) \in \mathbb R^2_+ \mid xy \ge 1\} $$ Consider the bounding hyperbola $xy=1$ (where $x,y > 0$). For each point $(a,b)$ on this hyperbola, first find the tangent line to the hyperbola at that point, and then find the half-space of that tangent line which contains the hyperbola. Your set $H$ is the intersection of those half-spaces, one for each such point $(a,b)$.
In detail, using calculus we find that the tangent line to $xy=1$ at $(a,b)$ has slope equal to $-\frac{1}{a^2}$, and therefore its equation is given in point slope by the formula $$\frac{y-b}{x-a} = - \frac{1}{a^2} $$ which can be rewritten $$x + a^2 y = a + a^2 b $$ We can eliminate $b=\frac{1}{a}$ to get $$x + a^2 y = 2a $$ The corresponding half-space which contains the hyperbola is $$H_a = \{(x,y) \in \mathbb R^2 \mid x + a^2 y \ge 2a\} \quad\text{for all $a > 0$} $$ Your set $H$ can then be written as the intersection of these half-spaces: $$H = \bigcap_{a > 0} H_a $$ Now there's something to prove: you must show that for all $(x,y) \in \mathbb R^2_+$ the following is true:
Even though this solution was discovered by calculus, I am guessing that this "if and only if" can be proved with straightforward algebra and inequalities, but maybe I'll stop here.