Quasi-concavity of A Minimized Quasi-concave Function

Question

Let function $F(x,y,z)$ defined on $[0,1]\times[0,1]\times[0,1]$ be increasing in $x$ and $y$. By increasing I mean $\frac{\partial F}{\partial x}\geq0$ and $\frac{\partial F}{\partial y}\geq0$.

By quasi-concave I mean $$F(x',y',z)\geq F(x,y,z)\Rightarrow[\frac{\partial F}{\partial x},\ \frac{\partial F}{\partial y}][x'-x,\ y'-y ]^T\geq0.$$

Define $$H(x,y)=\min\limits_{z\in[x,1]}F(x,y,z).$$

My question: Is function $H(x,y)$ quasi-concave in $x$ and $y$? Why?

EDIT 1: I think $H(x,y)$ is still increasing in $x$ and $y$, because for any $x'\geq x$ and $y'\geq y$, we have $\min\limits_{z\in[x',1]}F(x',y,z)\geq \min\limits_{z\in[x,1]}F(x,y,z)$ and $\min\limits_{z\in[x,1]}F(x,y',z)\geq \min\limits_{z\in[x,1]}F(x,y,z)$. Is it right?

EDIT 2: Correction of the quasi-concavity.

Answer 1

An increasing function is only necessarily quasiconcave if it is a map from $\mathbb{R}$ to $\mathbb{R}$. The function $x^2+y^2+z^2$ is strictly convex on $[0,1]^3$, and increasing for any meaningful definition of increasing.

Answer 2

Since, $F(x, y, z)$ is increasing, $$\ H(x, y) = \min_{z\in[x, 1]}F(x, y, z) = F(x, y, x) $$ Can you finish it now?