For $f(x,y) = \sqrt{x^2 + y^2 + \frac{1}{x^2 y^2}}$ the critical points are $(1,1), (1,-1), (-1,1), (-1,-1)$ and all of their values in the function is $3$. How can you tell that these are either maximums or minimums?
The example problem states that because the values of the critical points are all $3$, this implies that they are minimums without requiring any additional computations. How can you tell just by observation?
The auxiliary function $$g(u,v):=u+v+{1\over uv}\quad \bigl((u,v)\subset Q:=({\mathbb R}_{>0})^2\bigr)$$ is $\geq M$ when $u+v\geq M$ and goes to $\infty$ when $u\to0+$, or $v\to0+$. It follows that $g$ assumes a global minimum on $Q$ and that this minimum occurs at a critical point $(\xi,\eta)\in Q$. Computation shows that $(1,1)$ is the unique critical point of $g$ in $Q$, so that without further calculations we can say that $$g(u,v)\geq g(1,1)=3\quad\forall (u,v)\in Q\ .\tag{1}$$ The function $f$ defined in the question has as its domain the union of the four open quadrants of the $(x,y)$-plane, and is even with respect to $x$ and to $y$. It therefore suffices to study $f$ in the first quadrant $Q$. The substitution $u:=x^2$, $v:=y^2$ transforms $Q$ to $Q$ and $f^2$ into the function $g$ studied above. From $(1)$ we therefore can conclude that $$f(x,y)\geq f(1,1)=\sqrt{g(1,1)}=\sqrt{3}\qquad\bigl((x,y)\in Q\bigr)\ .$$ It follows that $\sqrt{3}$ is the global minimum value of $f$ on its domain.
Consider the square of the function and apply AM-GM inequality (arithmetic mean is greater than or equal to the geometric mean) and you are done !!
