Let
$$R := \{ (x,y) \in \mathbb{R}^2 :x \geq 0, y \geq 0, x^\alpha y^\beta \geq F_0 \}$$
where $\alpha > 0, \beta > 0$ and $F_0 > 0$. Is set $R$ convex?
We know that the intersection of convex set is convex. Now $\{x\geq 0, y\geq 0\}$ is clearly convex. So I need to show that the set $\{x^\alpha y^\beta \geq F_0\}$ is convex. I may show that the function $f(x,y) = x^\alpha y^\beta$ is concave, because this implies that for all $c \in \mathbb{R}$, the set $\{x : f(x) \geq c\}$ is convex. But I am not able to show that the function f is concave.
You actually only need to know that $$F(x,y)=x^{\alpha}y^{\beta}$$ is quasi-concave, i.e., that the superlevel sets $\{(x,y)\mid F(x,y)\geq F_0\}$ are convex. Concave functions are quasi-concave, and quasi-concave functions are preserved under monotone maps, so it suffices to show that $$\ln F(x,y)=\alpha\ln x + \beta \ln y$$ is a concave function. But that can be shown directly.