Proving $\{(x_1, x_2) \in \mathbb{R}^2_{+} \mid x_1 x_2 \geq 1\}$ is convex This is so called a hyperbolic set
$$\{(x_1, x_2) \in \mathbb{R}^2_{+} \mid x_1 x_2 \geq 1\}$$
We proceed to prove that it is convex by showing that a convex combination of points (a line segment) will lie in the set
Suppose $x = (x_1, x_2)$, $y = (y_1, y_2)$ and $x \geq y$ in the elementwise sense
Then set:
$z = \theta(x_1,x_2) + (1-\theta)(y_1, y_2)$
$z = (y_1,y_2) + \theta(x_1 - y_1, x_2 - y_2)$
Clearly, any $z_1,z_2 \in z$ lies in the hyperbolic set (i.e. $z_1z_2 \geq 1$)
How do we proceed to prove the case when $x \not \geq y$ in the elementwise sense?
 A: Say $f(x)=\frac{1}{x}$, so our set is $\{(x,y): y\geq f(x)\}$. We know that $f(x)$ is convex, that is for $0\leq\theta\leq 1$ $$f(\theta x+(1-\theta)y)\leq \theta f(x)+(1-\theta)f(y).$$
Now take $(x_1,y_1)$ and $(x_2,y_2)$ in the set. In other words $y_1\geq f(x_1)$ and $y_2\geq f(x_2)$. For $0\leq\theta\leq 1$
$$\theta y_1+(1-\theta)y_2\geq\theta f(x_1)+(1-\theta)f(x_2)\geq f(\theta x_1+(1-\theta)x_2)$$ so it is in the set.
A: Let $(x_1,y_1)$ and $(x_2, y_2)$ in our set, and $\alpha \in [0,1]$. We need to show that $\alpha (x_1,y_1) + (1-\alpha) (x_2, y_2)$ is in our set, i.e. that $$\alpha^2 x_1y_1 + (1-\alpha)^2 x_2y_2 + \alpha(1-\alpha)(x_1y_2+x_2y_1) \geq 1$$
Hint: notice that $x_1y_1 \geq 1, x_2y_2 \geq 1$ and prove that $x_1y_2 + x_2y_1 \geq 2$ to conclude.
A: Your hyperbolic set is defined by the following linear matrix inequality (LMI)
$$\begin{bmatrix} x_1 & 1\\ 1 & x_2\end{bmatrix} \succeq \mathrm O_2$$
which, using Sylvester's criterion, produces the following conjunction of (non-strict) inequalities
$$x_1 \geq 0 \land x_2 \geq 0 \land x_1 x_2 - 1 \geq 0$$
Hence, the hyperbolic set is a spectrahedron and, thus, convex.

convex-analysis linear-matrix-inequality spectrahedra
