<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] Let $x^2=\frac{2}{3}a$ and $y^2=\frac{3}{2}b$.
Hence, $ab=1$ and by C-S and AM-GM we obtain:
$$\frac{4}{4-x^2}+\frac{9}{9-y^2}=\frac{4}{4-\frac{2}{3}a}+\frac{9}{9-\frac{3}{2}b}=$$
$$=6\left(\frac{1}{6-a}+\frac{1}{6-b}\right)\geq\frac{6(1+1)^2}{6-a+6-b}=$$
$$=\frac{24}{12-a-b}\geq\frac{24}{12-2}=\frac{12}{5}.$$
The equality occurs for $a=b=1$, which says that $\frac{12}{5}$ is a minimal value.
Done!
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