There is a value less than Global minimum!

Question

Here we see that minimum of the function $x+y+1/xy$ is obtained at $(1,1)$ which is $3$

However , $f(1,-1) = -1$. Then how is $f(1,1)$ actually the minimum?

Answer 1

It's not! The key is local minimum, meaning that $f(1, 1)$ is less than $f(x, y)$ for all $(x, y)$ near $(1, 1)$. More precisely, there is some $r > 0$ such that $$\sqrt{(x - 1)^2 + (y - 1)^2} < r \implies f(x, y) \ge f(1, 1).$$ But, $f(x, y) \ge f(1, 1)$ doesn't hold globally, i.e. $f$ does not have a global minimum at $(1, 1)$.

While local minimums can usually be determined purely with derivatives (which themselves are purely about the local behaviour of functions), global minimums are more tricky. You usually have to consider the behaviour of $f$ as $(x, y)$ gets very large (Can it go to $-\infty$? If not, how small can it get?), as well as compare the function value at each critical point (or at least, the critical points which might be local minimums).

In this case, if we take $(x, y) = (-t, -t)$ as $t \to \infty$, note that $$f(x, y) = -2t + \frac{1}{t^2} \to -\infty.$$ That is, $f$ can become arbitrarily large and negative, and hence there will be no global minimum.