Prove that $\frac{x^2y}{1+x^4+y^2}$ has no global minimum I am not sure how to approach this problem. The usual methods do not work to find a minimum.
I can see that, but how to show that there must not exist a minimum?

$$f(x,y) = \frac{x^2y}{1+x^4+y^2}$$
$$\frac{\partial f}{\partial x} = \frac{2xy(1+x^4+y^2) - 4x^5y}{(1+x^4+y^2)^2}$$
$$\frac{\partial f}{\partial y} = \frac{x^2(1+x^4+y^2) - 2x^2y^2}{(1+x^4+y^2)^2}$$
$$\frac{\partial f}{\partial x} = 0 \implies 1+x^4+y^2 = 2x^4 \lor x=0 \lor y=0$$
$$\frac{\partial f}{\partial y} = 0 \implies 1+x^4+y^2 = 2y^2 \lor x=0$$
One check that when $x=0$ or $y=0$, the function is zero, so it is not the global minimum.
Otherwise, on solving the equations we have $y=x^2$, and $1+2x^4=2x^4 \implies 1 = 0$.
