I want to show that a function $\displaystyle \Psi (a_1,a_0)$, which is separable (additively decomposed) in two quasiconcave functions, is also quasiconcave (QC). I know that the sum of QC functions is not generally QC, but my numerical simulations seem to suggest that $\displaystyle \Psi (a_1,a_0)$ is.


I have tried using the definition (i.e. trying to show that it has convex upper contour sets by plugging $\displaystyle \Psi\big(\lambda a + (1-\lambda)b\big)\geq$... etc), but I can not prove the inequality. I also tried using the border Hessian, but i cannot sign the determinant.
Any other ideas??


Below is a more detailed description of the problem:


$\displaystyle \Psi (a_1,a_0)=S(a_1)+S(a_0)$


Where $\displaystyle a_1\in [0,1] $ and $\displaystyle a_0 \in [0,1]$, and


$\displaystyle S(a_j)=F\big(Bh(a_j)\big) \int_{a_j}^{1} \big(t*P(t)\big)dt + F\big(Bl(a_j)\big)\int_{a_j}^{1} \big(t*(1-P(t))\big)dt$


for $\displaystyle j \in{1,0}$


and the functions F(.), Bl(.), Bh(.) and P(.) are strictly increasing. Additionally, Bl''>0, and Bh''<0.


F'' and P'' can be positive or negative (as required to guarantee quasiconcavity --my simulations suggest that only P''>0 is necessary).


I would appreciate any hints --including other forums where I could post this question