高校数学(1) 根号と分数式の混じった式の最小値

Find the value of next expression.

$$\begin{array}{r}\underset{x\in (0,1)}{min}\{{\displaystyle \frac{1}{x^4}}+{\displaystyle \frac{x^2}{\sqrt{1-x^4}}}\}\end{array}$$

(Answer)

$$\begin{array}{rl}& \mathrm{Using}\mathrm{AM}-\mathrm{GM}\mathrm{inequality},\mathrm{we}\mathrm{get}\\ & {\displaystyle \frac{1}{x^4}}+{\displaystyle \frac{x^2}{\sqrt{1-x^4}}}=1+{\displaystyle \frac{1-x^4}{x^4}}+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{x^2}{\sqrt{1-x^4}}}+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{x^2}{\sqrt{1-x^4}}}\\ & \geqq 1+3\sqrt[3]{{\displaystyle \frac{1-x^4}{x^4}}\cdot {\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{x^2}{\sqrt{1-x^4}}}\cdot {\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{x^2}{\sqrt{1-x^4}}}}=1+{\displaystyle \frac{3}{2^{2/3}}}\\ & (\mathrm{equal}\mathrm{sign}\iff {\displaystyle \frac{1-x^4}{x^4}}={\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{x^2}{\sqrt{1-x^4}}}\iff x={\displaystyle \frac{2^{1/6}}{\sqrt[4]{1+2^{2/3}}}}\in (0,1))\\ & \mathrm{Hence},\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{expression}\mathrm{is}\underset{―}{1+{\displaystyle \frac{3}{2^{2/3}}}}\end{array}$$

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