Is this set convex ?2 Is this set convex for every arbitrary $\alpha\in \mathbb R$?
$$\Big\{(x_1,x_2)\in \mathbb R^2_{++} \,\Big|\, x_1x_2\geq \alpha\Big\}$$ 
Where $\mathbb R^2_{++}=[0,+\infty)\times [0,+\infty)$.
 A: For $\alpha\le 0$ your set equals $\mathbb R^2_{++}$, which is of course convex. For $\alpha>0$ you have $x_1,x_2\neq 0$ for all $(x_1,x_2)$ in the set and you can rewrite $x_1 x_2\ge \alpha$ as $x_2 \ge \alpha/x_1$, so your set is the region above the hyperbola $x_2=\alpha/x_1$ in $\mathbb R^2_{++}$, which is convex.
A: Yes, it is. You should separate the case $a<0$ (when it is the quadrant) and $a>0$ when it is bounded by an arc of a hyperbola.
A: I'm going to relax this slightly to $x_1,x_2\geq 0$. The $\alpha< 0$ case is trivial, as others have shown. Here's another way to handle the case when $\alpha\geq 0$. First, note that
$$
x_1x_2 \geq \alpha \quad\Longleftrightarrow\quad \tfrac{1}{4}(x_1+x_2)^2\geq \tfrac{1}{4}(x_1-x_2)^2+\alpha
$$
You'll want to multiply out the quadratics on the right-hand side to confirm this for yourself. Take the square root of both sides:
$$
\tfrac{1}{2}|x_1+x_2|\geq \sqrt{\tfrac{1}{4}(x_1-x_2)^2+\alpha} = \left\| \begin{bmatrix} \tfrac{1}{2}(x_1-x_2) \\ \sqrt{|\alpha|} \end{bmatrix} \right\|_2
$$
But of course, under our assumptions, $|x_1+x_2|=x_1+x_2$ and $|\alpha|=\alpha$. So putting it together,
$$
x_1,x_2\geq 0,~x_1x_2 \geq \alpha \quad\Longleftrightarrow\quad \left\| \begin{bmatrix} \tfrac{1}{2}(x_1-x_2) \\ \sqrt{\alpha} \end{bmatrix} \right\|_2 \leq \tfrac{1}{2}(x_1+x_2)
$$
which represents the composition of a convex second-order (Lorentz) cone and an affine mapping $(x_1,x_2)\rightarrow(\tfrac{1}{2}(x_1+x_2),\tfrac{1}{2}(x_1-x_2),\alpha)$.
This may seem like an awfully convoluted way to go about it. But in fact this is a common transformation in practice, because most primal-dual convex optimization solvers cannot handle the hyperbolic form directly, while the Lorentz cone is a fundamental object.
