<s>[INST] minimum value of a two variable problem Suppose that $x$ and $y$ are in $(−2, 2)$ and $xy = −1$. The minimum value of 
$\frac4{4-x^2}+\frac9{9-y^2}$ is ?
So I have tried to manipulate this equation but what I got is a complex equation with square roots in it...
 [/INST] $x, y \in (−2, 2)$ and $xy = −1$.
Let $3x+2y=k$. Then $9x^2-12+4y^2=k^2$. That is, $9x^2+4y^2=k^2+12$.
\begin{align}
   f(x,y)
   &= \frac4{(4-x^2)}+\frac9{(9-y^2)} \\
   &= \dfrac{72 - (9x^2+4y^2)}{36-(9x^2+4y^2)+1} \\
   &= \dfrac{72 - (k^2+12)}{36-(k^2+12)+1} \\
   &= \dfrac{60 - k^2}{25-k^2} \\
   &= 1 + \dfrac{35}{25-k^2}
\end{align}
This attains its smallest value, $\frac{12}{5}$, when $k=0$, for example, when $x = \frac 23$ and $y=-1$.
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