I'm stuck on the following proof. I want to show that $f(x_1,x_2) = x_1x_2$ on $\mathbf{R}_{++}^2$ is quasi-concave, i.e., all of its superlevels are convex.
$Proof.$ Take $(y_1,y_2)$ and $(z_1,z_2)$ from an arbitrary superlevel of $f$, say $\{(x_1,x_2)\in \mathbf{R}_{++}^2|x_1x_2\geq \alpha\}$. It suffices to show that $\theta(y_1,y_2) + \bar{\theta}(z_1,z_2)$ belongs to the superlevel where $0 \leq \theta \leq 1$ and $\bar{\theta}=1-\theta$. We have
$(\theta y_1+\bar{\theta}z_1)(\theta y_2+\bar{\theta}z_2) = \\ \theta^2y_1y_2 + \theta\bar{\theta}y_1z_2 + \theta\bar{\theta}y_2z_1 + \bar{\theta}^2z_1z_2 \\ \geq \theta^2\alpha + \theta\bar{\theta}(y_1z_2 + y_2z_1) + \bar{\theta}^2\alpha \\ \geq \theta^2\alpha + \bar{\theta}^2\alpha $
and then I'm stuck.
The problem is you don't want to ignore the cross terms. We have $$ y_1z_2 + y_2z_1 \geq 2\sqrt{y_1z_2y_2z_1}\geq 2\alpha.$$ Then continuing with your bound, we see that the lower bound is $\theta^2\alpha + 2\theta\bar{\theta}\alpha + \bar{\theta}^2\alpha = \alpha$.