<s>[INST] Minimum value of $\frac{4}{4-x^2}+\frac{9}{9-y^2}$ Given $x,y \in (-2,2)$ and $xy=-1$
Minimum value of $$f(x,y)=\frac{4}{4-x^2}+\frac{9}{9-y^2}$$
My try:
Converting the function into single variable we get:
$$g(x)=\frac{4}{4-x^2}+\frac{9x^2}{9x^2-1}$$
$$g(x)=\frac{4}{4-x^2}+1+\frac{1}{9x^2-1}$$
Using Differentiation we get:
$$g'(x)=\frac{8x}{(4-x^2)^2}-\frac{18x}{(9x^2-1)^2}$$
$$g'(x)=2x\left(\frac{4(9x^2-1)^2-9(4-x^2)^2}{(4-x^2)^2(9x^2-1)^2}\right)$$
$$g'(x)=70x\frac{9x^4-4}{(4-x^2)^2(9x^2-1)^2}=0$$
So the critical points are:
$x=0, x=\pm \sqrt{\frac{2}{3}}$
But $x \ne 0$ since $xy=-1$
$$g'(x)=70x \frac{(3x^2-2)(3x^2+2)}{(4-x^2)^2(9x^2-1)^2}$$
By using derivative test we get Minimum occurs when $x=\pm \sqrt{\frac{2}{3}}$
Hence $$x^2=\frac{2}{3}, y^2=\frac{3}{2}$$
Min value is $$\frac{4}{4-\frac{2}{3}}+\frac{9}{9-\frac{3}{2}}=\frac{12}{5}$$
Is there any other approach? 
 [/INST] From “harmonic mean $\le$ arithmetic mean” one gets
$$
\frac{2}{f(x, y)} \le \frac12 \left(\frac{4-x^2}{4}+ \frac{9-y^2}{9} \right)
= \frac 12 \left(2- \frac{x^2}{4} - \frac{y^2}{9} \right)
$$
and from “arithmetic mean $\ge$ geometric mean”
$$
  \frac{x^2}{4} + \frac{y^2}{9}\ge 2 \frac{|xy|}{6} = \frac 13
$$
so that
$$
f(x, y) \ge \frac{4}{2-\frac 13} = \frac{12}{5} \, .
$$
In both estimates equality holds if $\frac{x^2}{4} = \frac{y^2}{9} \iff x^2=\frac{2}{3}, y^2=\frac{3}{2},$ i.e. equality is attained the two points 
$$
  ( \sqrt \frac 23, -\sqrt \frac 32) \quad \text{and} \quad
  ( -\sqrt \frac 23, \sqrt \frac 32)
$$
which are both in the given domain $(-2, 2) \times (-2, 2)$ of $f$.
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