My assignment is to find the $min$ and $max$ (or $inf$ and $sup$) of the set $$\mathbb{A} := \left\lbrace \frac{|1-xy|}{(1+x^2)(1+y^2)}:\ x,y \in \mathbb{R}, x\neq y\right\rbrace$$
I know that $\inf \mathbb{A} = \min \mathbb{A} = 0$ and $\sup \mathbb{A} = 1$. The assignments says there is no explanation required. I found these by "guessing" However, I want to know how to find these two values in a mathematical way.
First of all, using $2xy \le x^2 + y^2$, $$|1-xy| \le 1+|xy| \le 1+ \frac{x^2}{2} +\frac{y^2}{2} \le 1+x^2 +y^2\le (1+x^2)(1+y^2),$$ so the function value is less than one. The above equality holds only when $x = y=0$, so the function value is strictly less then one (as $x\neq y$ by requirement). On the other hand, let $y = 0$, it becomes $$\frac{1 + 0}{(1+x^2)(1+0)} = \frac{1}{1+x^2}.$$ Let $x\to 0$, one see that the supremum (not maximum) is one.
The minimum is easier: it's obviously non-negative and by finding some $x,y$ so that $xy=1$, the minimumu (thus also the infimum) has to be zero.