Proving that $\exp(-x^2-y^2)$ is a strictly quasi-concave function

Question

Prove that $$f(x,y)=\exp(-x^2-y^2)$$ is a strictly quasi-concave function, that is,$\forall (x_i,y_i)\in \Re^2, (x_i,y_i)\neq (x_j,y_j), \lambda\in(0,1)$, $$f(\lambda x_1+(1-\lambda) x_2,\lambda y_1+(1-\lambda) y_2)>\min[f(x_1,y_1),f(x_2,y_2)].$$

I'm trying to use triangle inequality but so far without success. Perhaps some insight will be useful.

Answer 1

The logarithm is a strictly increasing monotonic function, so if $a>b$ then $\log a>\log b$ and vice-versa. Take logarithms then $$ \log (e^{-x^2-y^2})=-2(\log x+\log y) $$ Using just one dimension, suppose that the maximum is at $x_1$ then $$ \log \exp -(\lambda x_1+(1-\lambda)x_2)^2= -2(\lambda x_1+(1-\lambda)x_2)\\ >-2x_1 $$ by the usual inequalities.

Answer 2

Another approach is to observe a function $f$ is (strictly) quasi-concave if and only if all the upper contours set $\{x: f(x)\ge \alpha\}$ are (strictly) convex.