How can I show that the function $u(x_1,x_2)=(x_1x_2)^\alpha$ is quasiconcave, given $\alpha>1,x_i\geq0$?
I managed to find the bordered Hessian, whereby
$(-1)^1B_1=\alpha^2(x_1x_2)^{2(\alpha-1)}x^2_2\geq0$
and
$(-1)^2B_2=0\geq0$.
Nonetheless, that only provides a necessary condition for quasiconcavity.
Is it possible to show quasiconcavity from its definition, i.e., $u(ax_1+(1-a)y_1,ax_2+(1-a)y_2)\geq \textrm{min}\{u(x_1,x_2),u(y_1,y_2)\}$?
Is it possible to show quasiconcavity from its definition, i.e., $u(ax_1+(1-a)y_1,ax_2+(1-a)y_2)\geq \textrm{min}\{u(x_1,x_2),u(y_1,y_2)\}$?
Answer: Yes.
A useful trick that can save you some trouble is to perform a monotonic transformation. In preference relation terms you are trying to show
$$ \left( ax_1+(1-a)y_1,ax_2+(1-a)y_2 \right) \succeq \left[ (x_1,x_2) \text{ OR } (y_1,y_2) \right]. $$
If this holds for the utility representation $(x_1x_2)^{\alpha}$, it will also hold for monotonic transformations of this function (the ordering of baskets is unchanged).
Clearly there is a more elegant function to represent the relation, making mathematical calculations easier.
Alternatively you can power through it and make some assumptions w.o.l., like $u(x_1,x_2) \leq u(y_1,y_2)$. The function is clearly strictly monotonic, so that saves you from looking at cases where $x_1 \leq y_1$ AND $x_2 \leq y_2$. All that remains to check (w.o.l.) is the case where $x_1 < y_1$, $x_2 > y_2$.
