Assuming $f: \mathbb{R}^2 \rightarrow \mathbb{R}, x \in [0,+\infty], \alpha \in [0,1]$
I'm trying to prove that any upper level of the function $f(x) = x_1 ^ \alpha \cdot x_2 ^ {1-\alpha}$ is convex when $f(x)=c, \forall c \in \mathbb{R}$.
I know a function is quasi-concave if and only if a given set $X \in \mathbb{R}$ is convex, $\forall x \in X : f(x) \geq c$.
I remember that a function is convex if any line segment between two points inside the set is also part of the set. Therefore, $\forall p,q \in X, \lambda \in [0,1]$, if I define $z = \lambda p + (1-\lambda)q$, then the function is convex if $z \in X$.
I've tried developing on it but I don't know how to continue. It would be great if someone could help me!
If $c=0$ the proposition is not true, if $c\gt 0$, to facilitate computations let's assume $c=1$, then $f(x)=1 \implies x_1=x_2^{1-1/\alpha}$, then it remains to prove that the set $S=\{(x_1,x_2) | x_1\geq x_2^{1-1/\alpha}\}$ is a convex set.
Let $(a,b),(c,d)\in S$, then $a\geq b^{1-1/\alpha}$ and $c\geq d^{1-1/\alpha}$, it remains to prove that $\forall t \in [0,1]$ the point $e_t=t(a,b) +(1-t)(c,d)$ belongs to $S$.
Using the relations above and the fact that the function $g(x)=x^{1-1/\alpha}$ is convex (and this implies that $g(x+y)\leq g(x)+g(y)$ for every $x,y$) should be enough to prove the thesis.