Why $f:R^2\to R$, with $\mbox{dom} f=R^2_+$ and $f(x_1,x_2) =x_1x_2$ is quasiconcave? Why $f:R^2\to R$, with $\mbox{dom} f=R^2_+$ and $f(x_1,x_2) =x_1x_2$ is quasiconcave?
I have tried to use Jensen eniquality to check that superlevel set $\{x\in R^2_+ | x_1x_2 \ge \alpha\}$ is convex.
$$\begin{align*}(\theta x_1 + (1-\theta) x_3)(\theta x_2 + (1-\theta) x_4) \\= \theta^2 x_1x_2 + (1-\theta)^2x_3x_4+\theta(1-\theta)x_1x_4+\theta(1-\theta)x_2x_3\end{align*}$$
this must be greater or equal to 
$$\alpha$$
But here i don't know how to prove it.

Let $(x_1,y_1),\, (x_2,y_2) \in \{x,y\in\mathbb{R}_+: (x,y)\ge \alpha\}$. Then $y_1 \ge {\alpha \over x_1}$ and $y_2 \ge {\alpha \over x_2}$. Since ${1\over x}$ is convex. $${\alpha \over \theta x_1 + (1-\theta)x_2} \le \theta {\alpha \over x_1} + (1-\theta){\alpha \over x_2}$$ Then $${\alpha \over \theta x_1 + (1-\theta)x_2} \le \theta y_1 + (1-\theta)y_2 \implies \alpha \le (\theta x_1 + (1-\theta)x_2)\cdot(\theta y_1 + (1-\theta)y_2) = \theta^2 x_1 y_1 + \theta(1-\theta)(x_2 y_1+x_1 y_2) + (1-\theta)^2 x_2 y_2 = (\theta x_1 + (1-\theta)x_2)(\theta y_1 + (1-\theta)y_2)$$ and that means that $\{x,y\in\mathbb{R}_+: (x,y)\ge \alpha\}$ is convex.
