How to prove that a utility function U(x,y)=min(x,2y) is quasiconcave?
A function $f:D\rightarrow \mathbb{R}$ is said to be quasiconcave if the following set is a convex set for every value of $a\in\mathbb{R}$:
$P_a = \{x\in D: f(x) \geq a\}$
To show that $f(x,y) =\min(x, 2y)$ is quasiconcave, we just need to show that $P_a = \{(x,y)\in \mathbb{R}^2: \min(x, 2y) \geq a\}$ is a convex set. For that we consider arbitrary $(x', y')$ and $(x'', y'')$ from the set $P_a$ and arbitrary $\lambda\in [0,1]$ and show that
$\lambda (x', y')+(1-\lambda)(x'', y'')$ is in $P_a$. Observe that $x'\geq \min(x', 2y')\geq a$ and $x''\geq \min(x'', 2y'')\geq a$, so $\lambda x'+(1-\lambda)x''\geq a$. Likewise, $\lambda 2y'+(1-\lambda)2y''\geq a$. Therefore, it follows that $\min(\lambda x'+(1-\lambda)x'',2(\lambda y'+(1-\lambda)y'')) \geq a$ and consequently, $\lambda (x',y')+(1-\lambda)(x'', y'')$ is in $P_a$.