Prove that a function is quasi-concave Let $g^1\colon\mathbb{R}\to\mathbb{R}$ and $g^2\colon\mathbb{R}\to\mathbb{R}$ be concave functions, and let $f\colon\mathbb{R}\to\mathbb{R}$ be a non-decreasing function (i.e., $f(x)≥f(y)$ whenever $x≥y$).
Let $h\colon\mathbb{R}^2\to\mathbb{R}$ be defined by:
$$h(x_1,x_2)=f(g^1(x_1)+g^2(x_2)).$$
How do I prove that $h$ is quasi-concave?
 A: Note that if $\{x:g(x)>\alpha\}$ is convex for all $\alpha\in\mathbb{R}$, then $\{x:g(x)\ge\alpha\}=\bigcap\limits_{\beta<\alpha}\{x:g(x)>\beta\}$ is also convex.
Since $f$ is non-decreasing, $h$ is quasi-concave if $g^1(x_1)+g^2(x_2)$ is quasi-concave; i.e. either
$$
\small\{(x_1,x_2):f(g^1(x_1)+g^2(x_2))>\alpha\}=\{(x_1,x_2):g^1(x_1)+g^2(x_2)>\inf\{x:f(x)>\alpha\}\}\tag{1}
$$
or
$$
\small\{(x_1,x_2):f(g^1(x_1)+g^2(x_2))>\alpha\}=\{(x_1,x_2):g^1(x_1)+g^2(x_2)\ge\inf\{x:f(x)>\alpha\}\}\tag{2}
$$
where $(1)$ holds when $f(\inf\{x:f(x)>\alpha\})\le\alpha$ and $(2)$ holds otherwise.
Since both $g_1(x_1,x_2)=g^1(x_1)$ and $g_2(x_1,x_2)=g^2(x_2)$ are concave from $\mathbb{R}^2\to\mathbb{R}$, $g^1(x_1)+g^2(x_2)$ is also concave, hence quasi-concave.
A: There is an alternative. A function is quasi-concave if all super-level sets of it are convex. A super-level set for a function $z(x)$ is defined as $S_\alpha(z) = \{ x\, | \,z(x) > \alpha\}.$
Therefore we consider
 $$S_\alpha(h) =  \{ (x_1,x_2)\, | \,h(x_1,x_2) > \alpha\} = \{(x_1,x_2)\, |\, f(g^1(x_1)+g^2(x_2)) > \alpha\}$$
and since $f$ is a non-decreasing function so we have $$S_\alpha(h) = \{(x_1,x_2)\, |\, g^1(x_1)+g^2(x_2) > \alpha\}$$
Since according to problem statement, $g^1$ and $g^2$ are concave functions, so sum of them is also concave which have the pleasant indication that $S_\alpha(h) = \{(x_1,x_2)\, |\, g^1(x_1)+g^2(x_2) > \alpha\}$ is a convex set and therefore $h$ is a quasi-concave function.
A: Although an old question, I thought this deserved a general answer for multi-dimensional cases (following Sundaram's book "A first course in optimization theory"). 
One can test for quasi-concavity, by analyzing the bordered Hessian of a function $f(x_1,x_2,...,x_n)$, if it is 
a) twice differentiable in all arguments and 
b) its support is convex.
Let $H_k(x)$ denote the bordered Hessian, which is the $(k+1)*(k+1)$ matrix defined as follows: 
$$H_k(x) = \begin{bmatrix} 
    0 & \frac{\partial f}{\partial x_1}(x) & \dots & \frac{\partial f}{\partial x_k}(x) \\
 \frac{\partial f}{\partial x_1}(x) & \frac{\partial^2 f}{\partial^2 x_1}(x) & \dots & \frac{\partial^2 f}{\partial x_1 \partial x_k}(x) \\
    \vdots & \vdots & \ddots & \vdots \\
\frac{\partial f}{\partial x_k}(x) & \frac{\partial^2 f}{\partial x_k \partial x_1}(x) & \dots & \frac{\partial^2 f}{\partial^2 x_k}(x) \\
    \end{bmatrix}$$
Then, if $(-1)^k |H_k(x)| > 0$ $\forall$ $k \in {1,...,n}$, then $f$ is quasi-concave. $|H_k(x)|$ hereby is the determinant of $H_k(x)$. Note that this equality is strict.
