Global extrema of $(x^2 + y^2)e^{-(x^2 + y^2)}$ Could anyone please check my solution to the following problem?

Problem: Let $f(x, y) = (x^2 + y^2)e^{-(x^2 + y^2)}$. Find global extrema of $f$ on $M = {\mathbf R}^2$.

Proposed solution: Taking partial derivatives of $f$, we conclude that critical points are $[0,0]$ and points of the unit circle $C = \{[x,y] \in {\mathbf R}^2:\ x^2 + y^2 = 1\big\}$.
We can reason immediately that the global minimum is attained at $[0,0]$ as the function is nonnegative. We observe that the value of $f$ on $C$ is $e^{-1}$.
To prove that $f$ attains global maximum on $C$, we let $r := x^2 + y^2$. We observe that for any two points $[x_1,y_1]$ and $[x_2,y_2]$, $r_1 = r_2$ (ie. the function is constant on circles). Now let $r \to \infty$. Then
$re^{-r} \to 0$. From the definition of limit, it follows that for any $\varepsilon > 0$, we find $\delta > 0$ such that
$$\forall r \in P(\infty, \delta) = ({1 \over \delta}, \infty): re^{-r} < \varepsilon.$$
Let $\varepsilon = (2e)^{-1}$. Then there's $\delta$ from the definition above and we know that for $r \in ({1 \over \delta}, \infty)$, the value of f is less than the value of f on $C$. Restricting ourselves to the compact set
$$C' = \big\{[x, y] \in {\mathbf R}^2:\ x^2 + y^2 \le {1 \over \delta}\big\},$$
we can now argue that $f$ on $C'$ is indeed maximized on $C$, as it is a continuous function on a compact set, it's value around the boundary is at most $(2e)^{-1}$ and all critical points have been considered.
Therefore, the maximum value of $f$ is attained on $C$ with respect to $M$ as well.
 A: Looks good. Note that you can also start by translating the problem into a one-dimensional problem by letting $g(r) := re^{-r}$ and finding the global maximum and minimum of $g$ on $[0, \infty)$ and then note that $f(x,y) = g(x^2 + y^2)$ and conclude.
A: $$e^x\ge1+x \quad \forall x\in\mathbb{R}$$
(this is well-known / can be established all sorts of ways. Note that equality holds $\iff x=0$) 
Take $x=d-1$ to see 
$$e^{d-1}\ge d\implies de^{-d}\le e^{-1}\implies(x^2+y^2)e^{-(x^2+y^2)}\le e^{-1}$$
Chase back the fact that the equality holds when $x=0\iff d=1 \iff x^2+y^2=1$ to see that this maximum is attained, and that it is attained only on the unit circle.
Minimum is trivial for the reasons you detail - the function is non-negative, and $0$ only at the origin, so the global minimum is $0$ and achieved at $(0,0)$.
A: Or you could just remark that the function is radial with value $f(x) = |x|^2 e^{-|x|^2}$. so if it has a maximum/minimum at $x_0$, it's on the whole circle of radius $|x_0|$.
So it suffice to study the function $h(t) = t^2 e^{-t^2}$.
Here you can just differentiate, $h'(t) = (2t - 2t^3) e^{-t^2}$, and this is equal to $0$ for $t=\pm 1$, and it's easy to check that it's a maximum.
Hence $f$ is maximal on all $x$ such that $|x| = 1$
A: This looks correct to me, but one can treat this a little more efficiently: If $f$ achieves a global extremum at $(x, y)$, then the map $g: [0, \infty) \to \Bbb R$ defined by $$g(r) := r^2 e^{-r^2}$$ achieves a global extremum at $\sqrt{x^2 + y^2}$ and vice versa. Since $g$ is differentiable, to determine the latter it's enough to find the value of $g$ at the solutions of $g'(x) = 0$ and the value $g(0)$ at the endpoint $x = 0$. As was observed in the question, we know that $g$ achieves a minimum at $x = 0$ because evaluating gives that $g(0) = 0$ and $g(r) \geq 0$ for all $r$ (as $g$ is a product of nonnegative functions), leaving us just to solve $g'(r) = 0$, evaluate $M_a := g(r_a)$ for each solution $r_a$, and determine the maximum of the values $M_a$.
