<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] Hint: $f(x,y) = f(x) = \dfrac{4}{4-x^2} +\dfrac{9x^2}{9x^2-1}$. Let $u = x^2$, then $0 \leq u < 4$, and you get a relatively nice rational function in $u$ that you can use derivative to solve.
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