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.
Best Answer
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}