Find maximum and minimum value of inverse function .

Question

We have to find maximum and minimum value of $x^2 +y^2$

My try

how can I proceed

Answer 1

Let $\lfloor xy \rfloor=n$, then

$$\frac{x^2+y^2}{1-x^2y^2} = \tan \frac{n\pi}{4}$$

You've checked that no solutions for $0\le xy \le 1$, hence the tangent value must be negative which is achievable when $n=3,7,\ldots, 4k-1, \ldots$

Now \begin{align} \frac{x^2+y^2}{1-x^2y^2} &= \tan \left( k\pi-\frac{\pi}{4} \right) \\ x^2+y^2 &= x^2y^2-1 \end{align}

The required minimum value occurs when $xy=3$, therefore

$$\fbox{$\min \{ x^2+y^2 \}=8$}$$

and there're no maxima.