How to find the lower bound of $f(x, y) = -2xy+x^2y^2+x^2$? $f(x, y) = -2xy+x^2y^2+x^2, x\in \mathbb{R}, y\in \mathbb{R}$, how to find its lower bound?
Here are my thoughts, I don't know if it is rigorous.
$f(x, y) = -2xy+x^2y^2+x^2=(xy-1)^2+x^2-1$, as $(xy-1)^2 \geq 0$, $x^2\geq0$, and they can not get to $0$ at the same time, therefore $f(x, y) > -1$.
Therefore, the lower bound of $f(x, y)$ is $-1$.
 A: Your proof is certainly a proof that $-1$ is a lower bound of the expression.
However, I assume you want the largest lower bound. If that is the case, then your proof is not sufficient. You haven't proved, for example, that some larger value, say, $-\frac12$, isn't also a lower bound.

Hint:
To actually prove $-1$ is the largest lower bound, think about what happens when $y$ is very big and $x=\frac1y$. What happens to the first squared expression? What happens to the second one?
A: You have proven that the function is always greater than -1, but usually by lower bound we mean the inf of the values the function can attain; so you still need to prove that the function can actually attain the value of -1 (in case of a minimum; or get arbitrarily close to it in case of the infimum)
To make an example in one variable; the function $$f(x) = \frac 1{x^2 }+ 1$$ is always greater than $0$, but you wouldn’t call $0$ its lower bound right? The function is also bounded by $1$, which it achieves in the limit (so $1$ is the lower bound of the function, and the infimum of the values that the function achieves )
A: As you stated the function is $f(x, y)=(xy-1)^2+x^2-1$, and $x^2\ge0$, and $(xy-1)^2\ge0$. so it has a minimum when both of the inequalities are at their smallest
now $x^2$ is minimised when $x=0$, and at $x=0$, $(xy-1)^2=(-1)^2=1$, therefore $f(0,y)=0+1-1=0$,
But looking at $(x,y)=(\frac{1}{2},1)$, therefore $f(\frac{1}{2},1)=(\frac{1}{2}-1)^2+\frac{1}{2}^2-1=\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}$ so minimising along $x=0$ won't give us the minima of the function
So maybe tying along the line where $(xy-1)^2$ is minimised (ie when $xy-1=0$) and seeing if it also allows you to minimises $x^2$
hopefully that helps
