How to show that there isn't an absolute minimum/maximum for a $f(x,y)$ function? I'm dealing with this function $f(x,y)=\frac{1}{(x-a)^2+y^2+4} +\frac{1}{(x+a)^2+y^2+4}$ where $a$ is some constant.
I have realized that it attains it's maximum value at $(x,y)=(0,0)$ ,and that there isn't an absolute minimum since the function will keep attaining smaller values as we plug in higher/smaller values for $(x,y)$ and $a$.
How does one show this formally by using Weierstrass theorem? 
Edit:
The Weierstrass theorem was not relevant because the domain of $f(x,y), D$ isn't bounded.
Now I have assumed that the function attains it's absolute minimum $f(x,y)=m$ at $(x,y)=(x_0,y_0)$, so
$m\leq f(x,y) ,\forall(x,y)\Rightarrow \lim_{(x,y)\to\infty} m\leq \lim_{(x,y)\to\infty} f(x,y) \Rightarrow m\leq0.$
Now i want to show that $f(x,y)$ is always positive,I need help in proving that the function attains it's maximum at $(0,0)$.
 A: The function $f$ is strictly positive for all $(x,y)\in{\mathbb R}^2$. On the other hand, letting $r:=\sqrt{x^2+y^2}$ we obviously have $\lim_{r\to\infty} f(x,y)=0$. It follows that $f$ assumes no global minimum on ${\mathbb R}^2$.
Let $\epsilon:={1\over2}f(0,0)={1\over a^2+4}>0$. Then there is an $R>0$ such that $f(x,y)\leq\epsilon$ when $r\geq R$. The function $f$ assumes a global maximum $\geq2\epsilon$ on the closed disc $B_R$, and this maximum is then also the global maximum of $f$ on ${\mathbb R}^2$. Furthermore we can assert that this maximum will be taken in an interior point of $B_R$, hence at a critical point of $f$.
Inspection of the  expression defining $f$ shows that at the maximum point we necessarily have $y=0$. It is therefore sufficient to consider the critical points of the function
$$g(x):=f(x,0)={1\over(x-a)^2+4}+{1\over(x+a)^2+4}\ .$$
The numerator of $g'(x)$ computes to
$$N(x)=-4x\bigl(x^4+2x^2(4+a^2)+(4+a^2)(4-3a^2)\bigr)\ .$$
Analyzing the expression on the RHS we find the following: When $a\leq2/\sqrt{3}$ then there is only the critical point $x_0=0$ of $g$, which then necessarily gives the global maximum of $f$ on ${\mathbb R}^2$. But when $a>2/\sqrt{3}$ there are three critical points $x_{\pm1}$ and $x_0$. The graph of $g$ has two local maxima at $x_{\pm1}$ and a local minimum at $x_0=0$. The global maximum of $f$ on ${\mathbb R}^2$ is then given by $f(x_{\pm1},0)$. (The maximal value of $f$ can be explicitly computed.)

