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 $\begin{aligned}&\color{white}=f(\theta (u,v)+(1-\theta)(z,t))\\&=f(\theta u+(1-\theta)z, \theta v+(1-\theta)t)\\&>\min( f(u,v), f(z,t) )\\&=f(z,t)\end{aligned}$
$g$ is quasiconcave then $\begin{aligned}&\color{white}=g(\theta (u,v)+(1-\theta)(z,t))\\&=g(\theta u+(1-\theta)z, \theta v+(1-\theta)t)\\&>\min( g(u,v), g(z,t) )\\&=g(z,t)\end{aligned}$
I assume that $m(x,y)=\min(f(x,y), g(x,y))$
I want to show that $\begin{aligned}&\color{white}=m(\theta (u,v)+(1-\theta)(z,t))\\&=m(\theta u+(1-\theta)z, \theta v+(1-\theta)t)\\&>\min( m(u,v), m(z,t) )\end{aligned}$
$$\begin{aligned}&\color{white}=m(\theta (u,v)+(1-\theta)(z,t))\\&=m(\theta u+(1-\theta)z, \theta v+(1-\theta)t)\\&=\min(f(\theta u+(1-\theta)z, \theta v+(1-\theta)t), g(\theta u+(1-\theta)z, \theta v+(1-\theta)t))\\&=\end{aligned}$$
And then, how can I proceed this proof?
Any helps will be appreciated. Thank you.
Hint
$$\min\{\min\{a,b\},\min\{c,d\}\}=\min\{\min\{a,c\},\min\{b,d\}\}$$