proving the superlevel set is convex (quasiconcave)

Question

I want to show that $f(x_1, x_2) = x_1x_2$ is quasiconcave. Let $$S_\alpha = \{x_1, x_2 \in \mathbf{R}_{++}\; | \; f(x_1, x_2) = x_1x_2 \ge \alpha \}$$ and $(a_1, b_1), (a_2, b_2) \in S_\alpha$. So $a_1b_1 \ge \alpha$ and $a_2b_2 \ge \alpha$. Then \begin{align*} f(\theta a_1 + (1-\theta)a_2, \theta b_1 + (1-\theta)b_2) = \underbrace{\theta^2 a_1 b_1}_{\ge \theta^2 \alpha} + (1 - \theta)\theta (b_2a_1 + b_1a_2) + \underbrace{(1-\theta)^2a_2b_2}_{\ge (1-\theta)^2 \alpha} \end{align*} I know the lower bounds of the first and third term, but I don't know how to get the lower bound of the second term.

Answer 1

For the second term, you can use the following inequality (known as the AM-GM inequality): $$x + y \ge 2\sqrt{xy} \ \forall x,y\ge 0,$$ which is true because $x + y - 2\sqrt{xy} = (\sqrt{x} - \sqrt{y})^2 \ge 0.$

Applying this inequality we obtain $a_1b_2 + a_2b_1 \ge 2\sqrt{a_1a_2b_1b_2} \ge 2\alpha$.