<s>[INST] Find the minimum value of $\frac{4}{4-x^2} + \frac{9}{9-y^2} $ Let $x, y ∈ (−2, 2)$ and $xy = −1$. Find the minimum value of $\frac{4}{4-x^2} + \frac{9}{9-y^2} $ ?
My Attempt
let $t=\frac{4}{4-x^2} + \frac{9}{9-y^2} $ , replacing $y$ by $- \frac{1}{x}$ we get $t=\frac{1}{1-(\frac{x}{2})^2} + \frac{1}{1-(\frac{1}{3x})^2} $ . Using AM-HM inequality we get $t(1-(\frac{x}{2})^2 + 1-(\frac{1}{3x})^2) \geq 2^{2}.$  let $ m =(1-(\frac{x}{2})^2 + 1-(\frac{1}{3x})^2)$ and using AM-GM inequality we get $m \leq 5/3$.
But from this point my inequality signs are getting mixed up. Am I on right track?
 [/INST] EDIT : The previous solution was utterly mistaken.
Let$$\frac{4}{4-x^2} + \frac{9}{9-y^2} =f(x,y)$$
$$f(x,y)=\frac{36}{36-9x^2}+\frac{36}{36-4y^2} \ge \frac{(6+6)^2}{72-9x^2-4y^2} (\because \text{Cauchy})\ge \frac{144}{60}=\frac{12}{5}(\because \text {AM-GM})$$
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