$min\{f(x,y),g(x,y)\}$ is also quasiconcave for $f(x,y)$ and $g(x,y)$ quasiconcave functions

Question

Assume that $f(x,y)$ and $g(x,y)$ are quasiconcave functions, then $min\{f(x,y),g(x,y)\}$ is also quasiconcave.

What I did

Assume that $\{(u,v), (z,t)\}$ are domain of f and g for $0<\theta<1$

$f$ is quasiconcave then $f(\theta (u,v)+(1-\theta)(z,t))=f(\theta u+(1-\theta)z, \theta v+(1-\theta)t)>\min( f(u,v), f(z,t) )=f(z,t)$

$g$ is quasiconcave then $g(\theta (u,v)+(1-\theta)(z,t))=g(\theta u+(1-\theta)z, \theta v+(1-\theta)t)>\min( g(u,v), g(z,t) )=g(z,t)$

I assume that $m(x,y)=min(f(x,y), g(x,y))$

I want to show that $m(\theta (u,v)+(1-\theta)(z,t))=m(\theta u+(1-\theta)z, \theta v+(1-\theta)t)>\min( m(u,v), m(z,t) )$

$$m(\theta (u,v)+(1-\theta)(z,t))=m(\theta u+(1-\theta)z, \theta v+(1-\theta)t)=min(f(\theta u+(1-\theta)z, \theta v+(1-\theta)t), g(\theta u+(1-\theta)z, \theta v+(1-\theta)t))=$$

And then, how can I proceed this proof?

Any helps will be appreciated. Thank you.

*duplicated question

Answer 1

To show that $\min(f(x,y), g(x,y))$, consider the upper level set $$P^a = \left\{(x,y) \in \mathbb{R}^2 | \min(f(x,y), g(x,y)) \geq a\right\}$$

We'll show that this is a convex set for every $a\in\mathbb{R}$.

\begin{eqnarray*} P^a & = & \left\{(x,y) \in \mathbb{R}^2 | \min(f(x,y), g(x,y)) \geq a\right\} \\ & = & \left\{(x,y) \in \mathbb{R}^2 | f(x,y) \geq a\right\} \cap \left\{(x,y) \in \mathbb{R}^2 | g(x,y) \geq a\right\}\end{eqnarray*} which is the intersection of two convex sets because $f$ and $g$ are both quasi-concave, and hence it is also convex. Therefore, $\min(f(x,y), g(x,y))$ is quasi-concave.

Answer 2

You don't need to prove anything!

$H(x,y) = \min\{g(x,y), f(x,y)\}$is either equal to $g(x,y)$ or to $f(x,y)$, which are both quasi-concave by definition!