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?

Using AM-GM repeatedly:
$$\frac4{4-x^2}+\frac9{9-y^2} \ge \frac2{\sqrt{(1-x^2/4)(1-y^2/9)}} \ge \frac4{2-(x^2/4+y^2/9)}$$
Further, again as $\dfrac{x^2}4+\dfrac{y^2}9 \ge \dfrac13|xy|=\dfrac13$, we have
$$\frac4{4-x^2}+\frac9{9-y^2} \ge \frac4{2-\frac13}=\frac{12}5$$
Equality and the constraint is satisfied when $x = \sqrt{\frac23}, y = -\sqrt{\frac32}$, so that is the minimum.
