Prove ${\{x \in \mathbb R_{>0}^2 \mid x_1x_2 \ge \alpha \}}$ is a convex set for positive x

Question

Show that ${\{ x \in \mathbb R_{> 0}^2 \mid x_1x_2 \ge \alpha\}}$ is a convex set.

Using Jensen's inequality, let $x_1x_2 \ge \alpha$, and $y_1y_2 \ge \alpha$. For all $0 \le \theta \le 1$,

$$ \begin{align} (\theta x_1 + (1-\theta)y_1)(\theta x_2 + (1-\theta)y_2) & = \theta^2x_1x_2 + (1-\theta)^2y_1y_2 + \theta(1-\theta)(x_1y_2+y_1x_2)\\ & \ge \theta^2\alpha + (1-\theta)^2\alpha + \theta(1-\theta)(x_1y_2+y_1x_2)\\ & \ge \theta^2\alpha + (1-\theta)^2\alpha \\ & = (2\theta^2-2\theta+1)\alpha \end{align} $$

However, I'm not sure how to proceed. The fact that $(2\theta^2-2\theta+1)$ is only $\ge 1/2$ means I cannot say for sure that

$$(2\theta^2-2\theta+1)\alpha \ge \alpha$$

Note that I wish to prove this without using graphical method if possible.

Answer 1

One thing you can do is rigourize the "graphical". If you don't want to do that, you need to be more careful when throwing away terms. Note that $\alpha = 0$ is trivial and otherwise by scaling in one dimension, you can assume that $\alpha = 1$. Secondly, note the following identity for any $t>0$, which follows by calculus $$ t+t^{-1} \ge 2.$$ Set $t=\frac{x_1}{y_1}$. Then $x_1y_2 + x_2y_1 > \frac{x_1}{y_1} + \frac{y_1}{x_1} = t + t^{-1} \ge2 $. This gives \begin{align} (\theta x_1 + (1-\theta)y_1)(\theta x_2 + (1-\theta)y_2) & \ge \theta^2 + (1-\theta)^2+ \theta(1-\theta)(x_1y_2+y_1x_2)\\ & \ge \theta^2 + (1-\theta)^2 + 2 \theta (1-\theta) \\ & = (\theta + (1-\theta))^2 = 1. \end{align}

Answer 2

An easier proof is that the set is the epigraph of $f(x) = 1/x$ on $\mathbb{R}_{++}$. Showing that $f$ is convex is trivial (via the second derivative), and the epigraph of a convex function is convex.