China Mathematical Competition (Shaanxi)

Solution I We have $$\begin{array}{rl}u& =\frac{4}{4-x^2}+\frac{9x^2}{9x^2-1}=1+\frac{35x^2}{-9x^4+37x^2-4}\\ & =1+\frac{35}{37-\left({\left(3x-\frac{2}{x}\right)}^2+12\right)}\end{array}$$ Since $x\in (-2,-\frac{1}{2})\cup (\frac{1}{2},2)$, so $u$ reaches the minimum value $\frac{12}{5}$ when $x=\pm \sqrt{\frac{2}{3}}$. Answer: D.

Solution II It is known from the conditions that $4-x^2>0$ and $9-y^2>0$. Then $$\begin{array}{rl}u& \ge 2\sqrt{\frac{4}{4-x^2}\cdot \frac{9}{9-y^2}}=\frac{12}{\sqrt{36-9x^2-4y^2+(xy{)}^2}}\\ & =\frac{12}{\sqrt{37-9x^2-4y^2}}\ge \frac{12}{\sqrt{37-2\sqrt{36(xy{)}^2}}}=\frac{12}{5}.\end{array}$$ Since $u$ is $\frac{12}{5}$ when $x=\sqrt{\frac{2}{3}}$ and $y=-\sqrt{\frac{3}{2}}$, so $u$ reaches the minimum. Answer: D.