An algebra problem by Syed Hamza Khalid

$\begin{aligned} X & = \frac 4{4-x^2} + \frac 9{9-y^2} \\ & = \frac {4(9-y^2)+9(4-x^2)}{(4-x^2)(9-y^2)} \\ & = \frac {72-4y^2-9x^2}{36-9x^2-4y^2 +\color{#3D99F6} x^2y^2} & \small \color{#3D99F6} \text{Note that }xy = - 1 \\ & = \frac {72-4y^2-9x^2}{37-9x^2-4y^2} \\ & = 1 + \frac {35}{37-9x^2-4y^2} \\ & = 1 + \frac {35}{37-(9x^2+ {\color{#3D99F6}12xy} + 4y^2) + \color{#3D99F6}(-12)} \\ & = 1 + \frac {35}{25-\color{#3D99F6}(3x + 2y)^2} & \small \color{#3D99F6} \text{Note that }(3x + 2y)^2 \ge 0 \\ \implies X & \ge 1 + \frac {35}{25} = \boxed{\dfrac {12}5} \end{aligned}$

And equality occurs when:

$\begin{aligned} 3x + 2y & = 0 \\ 3x & = - 2y \\ 3x^2 & = - 2xy = 2 \end{aligned}$

$\implies x = \pm \sqrt {\frac 23} \implies y = \mp \sqrt{\frac 32}$ . Note that $x, y \in (-2, 2)$ , therefore the solution is valid.