I want to check does this set is concave
$$\{ (x,y,z) : 0 < z < xy \}$$
- My solution: it isn't concave as For example: $A_{1} (1; 1; 0.1)$ and $A_{2} (-1; -1; 0.1)$. Both of this points in the set. But their linear combination with $\lambda = 0.5$ isn't in the set $A_{3} (0; 0; 0.1) =>$ set isn't concave
I'm trying to use general definition of concave set i.e. $$0 < z_{1} + (1-\alpha)z_{2} < (\alpha x_{1} + (1-\alpha)x_{2})(\alpha y_{1} + (1-\alpha)y_{2}) <=> \\ 0 < z_{1} + (1-\alpha)z_{2} < \alpha^{2} x_{1} y_{1} + \alpha(1-\alpha)x_{1}y_{2} + \alpha(1-\alpha)y_{1}x_{2} + (1-\alpha)^{2}x_{2}y_{2} <=> \\ 0 < z_{1} + (1-\alpha)z_{2} < \alpha (\alpha x_{1} y_{1} + y_{1} x_{2} - \alpha y_{1} x_{2}) + (1-\alpha)(\alpha x_{1} y_{2} + x_{2} y_{2} - \alpha x_{2} y_{2})$$ $=>$ $\begin{cases} \alpha x_{1} y_{1} + (1-\alpha)x_{1}y_{2} > z_{1} \\ \alpha y_{1} x_{2} + (1-\alpha)x_{2} y_{2} > z_{2} \end{cases} <=>$ $\begin{cases} x_{1} (\alpha y_{1} + (1-\alpha)y_{2} > z_{1} \\ x_{2} (\alpha y_{1} + (1-\alpha)y_{2}) > z_{2} \end{cases}$
So, I don't know what to do next? I feel I'm pretty close to prove that it isn't concave set. But could you help me to solve it a little bit?