Finding critical points (extrema) directly without using eigenvalues If we have a function $f: \mathbb R^3\rightarrow \mathbb R$ of the form:
$$f(x)=(a_1x_1^2+a_2x_2^2+a_3x_3^2)e^{-x_1^2-x_2^2-x_3^2}$$
where $a_1>a_2>a_3>0$. 
We can find the critical points by:
$$\frac{\partial f}{\partial x_1} =2x_1e^{-x_1^2-x_2^2-x_3^2}(a_1-a_1x_1^2-a_2x_2^2-a_3x_3^2)=0$$
And solving symmetrically for $x_2$ and $x_3$. We get critical points: $(0,0,0) , (\pm 1,0,0) , (0,\pm 1, 0) , (0,0,\pm 1)$ 
We could determine the nature of them by looking at the eigenvalues of the Hessian, but is there a way to just analyse it directly? Obviously $(0,0,0)$ is a strict global minimum, but I can't quite see that the other points are. Any pointers/hints would be appreciated.
 A: We have $f(x,y,z) \geq 0$ and $f(0,0,0)=0$ so $(0,0,0)$ is a global minimum. The other critical points give rise to the function values
$$\left\{\frac{a_1}{e}, \frac{a_2}{e}, \frac{a_3}{e}\right\}$$
Since $f$ is continuous and bounded on $\mathbb{R}^3$ and approaches $0$ as $||x||\to \infty$ it has a maximum point(s). Since $\frac{a_1}{e} > \frac{a_2}{e} > \frac{a_3}{e}$ we must have that $(\pm 1,0,0)$ is a global maximum.

Close to $(0,\pm 1, 0)$ we have
$$f(\epsilon,\pm 1,0) = \left(\frac{a_2}{e} + \frac{a_1}{e}\epsilon^2\right)e^{-\epsilon^2} \approx \frac{a_2}{e} + \frac{a_1-a_2}{e}\epsilon^2 > \frac{a_2}{e}$$
and
$$f(0,\pm 1,\epsilon) = \left(\frac{a_2}{e} + \frac{a_3}{e}\epsilon^2\right)e^{-\epsilon^2} \approx \frac{a_2}{e} + \frac{a_3-a_2}{e}\epsilon^2 < \frac{a_2}{e}$$
so $(0,\pm 1, 0)$ are saddle points.

Close to $(0,0,\pm 1)$ we have
$$f(\epsilon,0,\pm 1) = \left(\frac{a_3}{e} + \frac{a_1}{e}\epsilon^2\right)e^{-\epsilon^2} \approx \frac{a_3}{e} + \frac{a_1-a_3}{e}\epsilon^2 > \frac{a_3}{e}$$
and since $x=1$ is a maximum for $x^2e^{-x^2}$ we have
$$f(0,0,\pm 1 \pm \epsilon) = a_3(1+\epsilon)^2e^{-(1+\epsilon)^2} < \frac{a_3}{e}$$
so $(0,0,\pm 1)$ are saddle points.
A: Other than $(0,0,0)$, which as you point out is a strict global minimum, the other critical points all lie on the unit sphere $S^2 \subset \mathbb{R}^3$. Restricted to that sphere, the exponential factor is just a constant $e^{-1}$. The critical points of the restricted function $g: S^2 \to \mathbb{R}$, $g(x_1,x_2,x_3) = \frac{1}{e}(a_1x_1^2 + a_2x_2^2 + a_3x_3^2)$ should be pretty easy to classify. Then you remember that all of those points are at a maximum with respect to the radial dimension.
