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?

Since $(x^2-|y|)^2 = x^4+y^2 - 2 x^2|y| \ge 0$, we have $x^4+y^2 \ge 2 x^2 |y|$.
Hence ${x^2|y| \over 1+x^4+y^2} \le {x^4+y^2 \over 1+x^2 + y}<{1 \over 2}$.
If we let $x(t) = \sqrt{|t|}, y(t) = -t$, we have $\lim_{t \to\infty } {-t^2 \over 1+ 2 t^2} = -{1\over 2}$.
In particular, $\inf_{x,y} {x^2 y \over 1+x^4+y^2} = - {1 \over 2}$, but
the value of $-{1 \over 2}$ is never attained.
