Finding the minimum of function including a $-\frac{1}{x}$ term.

Question

I would like to find the global minimum of the function: $$f(x, y) = 10(y^2-2x^3)^2 + (1-x)^2 - (y-1000)^{-1}.$$

Now my problem is the following. I know that $(1, \sqrt{2})$ is a local minimum, and I could try to prove it is also a global minimum. However, when investigating the function, I notice a term $-(y - 1000)^{-1}$ and I'm confused how to deal with this.

My Thoughts

Now, I thought the limit $\lim_{x \to 0} \frac{1}{x}$ didn't exist. However, to me it seems that $\lim_{x \downarrow 0} \frac{1}{x} = \infty$ and thus also for the function $f(x, y)$ I can say that $\lim_{y \downarrow 1000} \frac{-1}{y - 1000} = -\infty$? And thus the point $(1, \sqrt{2})$ can only be a local minimum?

Answer 1

The point $(1,\sqrt{2})$ is not a global minimum, because

$$f(1,\sqrt{2}) > f(1, 1000.000000001)$$

I don't really know the value of any of the two terms, but the simple fact that $(1000+\epsilon - 1000)^{-1} = \frac{1}{\epsilon}$ is enough for me to know that if I pick a small enough positive $\epsilon$, the expression $f(1, 1000+\epsilon)$ will be as small as I want.

Answer 2

We have $f(0,y) \to - \infty$ as $y \to 1000+$, hence $f$ does not have a global minimum.