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?

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Prove that for all $x_1,x_2,x_1',x_2'$ and $0\leq\alpha\leq 1$: $h(\alpha(x_1,x_2)+(1-\alpha)(x_1',x_2'))\geq\min \{f(x_1,x_2),f(x_1',x_2')\}$. – Davide Giraudo Oct 9 '11 at 7:30

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.

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