Show that there is no global minimum I want to show that the multivariablefunction $$f(x,y)=2x^4+5y^4-|x|-\sqrt{|x|+|y|}$$ has no global minimum.
For that do we calculate the critical points to get the desired result?
Or do we suppose that there is a global minimum at a point $(a,b)$ and then we have to show that there is a point $(x,y)$ with $f(x,y)\leq f(a,b)$ ?
 A: Note that $f(-x,y)=f(x,-y)=f(x,y)$, so we can focus on the positive quadrant $x,y\ge0$ and remove the absolute value signs:
$$f(x,y)=2x^4+5y^4-x-\sqrt{x+y}\qquad x,y\ge0$$
In any direction the quartic terms dominate, so there is no global maximum. On the other hand, $f$ in the positive quadrant is a sum of continuous convex functions ($2x^4,5y^4,-x,-\sqrt{x+y}$) and hence is continuous strictly convex, so there is a unique (up to sign) global minimum. An exact minimisation performed by Mathematica gives the global minimum in the positive quadrant as $$f(0.576690\dots,0.298945\dots)=-1.25130\dots$$
A: You can argue using the properties of continuous functions - as $x,y \to \infty$, $f \to \infty$, hence there is a closed ball $b(r) = \{ |(x,y)| \leq r\}$ such that if $x \in \mathbb R^2 \setminus b(r)$ then
$$
f(x,y) \geq  f(0,0)
$$
Now consider the restriction of $f$ to $b(r)$. Since $f$ is continuous and $b(r)$ is compact, the image $f(b(r))$ is also compact and hence the restriction of $f$ is bounded and attains its bounds. In particular there is a point $(x,y)$ such that  $f(x,y) \leq f(x',y')$ for all $(x',y') \in b(r)$. If $(x',y') \in \mathbb R^2 \setminus b(r)$ then
$$
f(x,y) \leq f(0,0) \leq f(x',y')
$$
Hence $(x,y)$ is a global minimum.
