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Consider the following function,

$$ f(x, y) = e^{m e^{-y}+n e^{-x}-x-y} \left(a x e^y+b e^x y+c x y\right) $$

where $a, b, c, m$ and $n$ are positive constants.

I want to show $f(x, y)$ is quasi-concave. To that end, we define the set $S_{\alpha} \subset \mathbb R^2$ by,

$$ S_{\alpha} = \{ (x, y) | f(x, y) > \alpha \} $$

How can I show the set $S_{\alpha}$ is convex? Is there any simpler way than using the definition $$f(x_1, y_1) > \alpha \wedge f(x_2, y_2) > \alpha \Rightarrow f(t x_1+(1-t)x_2, t y_1+(1-t)y_2) > \alpha \;\text{ for }\; 0 < t < 1?$$ Is it possible to show convexity by composing some basic functions like $xy$, $x e^y$, ...?

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I am not sure on how to go for the quasi-concave part, but my first guess would be to use derivatives to show your set's convex, would make your life a little easier. –  Patrick Da Silva Jun 2 '11 at 5:26
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I have no idea how can one show the convexity of a set by derivation. Any hint? –  Mohsen Jun 2 '11 at 6:15
    
@Mohsen The way of showing that $f(x,y)$ is convex by using partial derivatives means you will have to compose a Hessian matrix of $f(x,y)$. But it does not look like it is the most convenient method for this particular function to compute its partial derivatives. –  Koba Apr 28 '12 at 5:15

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