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?
 A: 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.
A: EDIT : The previous solution was utterly mistaken.
Let$$\frac{4}{4-x^2} + \frac{9}{9-y^2} =f(x,y)$$
$$f(x,y)=\frac{36}{36-9x^2}+\frac{36}{36-4y^2} \ge \frac{(6+6)^2}{72-9x^2-4y^2} (\because \text{Cauchy})\ge \frac{144}{60}=\frac{12}{5}(\because \text {AM-GM})$$
A: 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.
A: here is another elementary method which can also find min.
$f=\dfrac{4}{4-u}+\dfrac{9u}{9u-1},u=x^2=\dfrac{1}{y^2}< 4, y^2 < 4 \implies u> \dfrac{1}{4}$
$9(f-1)u^2+(72-37f)u+4(f-1)=0, \Delta=(72-37f)^2-4*4*9(f-1)^2 \ge 0 $
$\iff (12-5f)(12-7f)\ge 0 \iff f\ge \dfrac{12}{5}$ or $f \le \dfrac{12}{7}$
when $f \le \dfrac{12}{7}, 72-37f>0 \implies u<0$
when $f=\dfrac{12}{5}, u= \dfrac{37f-72}{2*9(f-1)}=\dfrac{2}{3} > \dfrac{1}{4}  \implies f_{min}=\dfrac{12}{5}$ 
