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{align} \min_{x\in(0,1)}\left\{\dfrac{1}{x^4}+\dfrac{x^2}{\sqrt{1-x^4}}\right\}\\ \end{align}

(Answer)

\begin{align} \tiny &\mathrm{Using\ AM-GM\ inequality,}\ \mathrm{we\ get}\\ &\dfrac{1}{x^4}+\dfrac{x^2}{\sqrt{1-x^4}}=1+\dfrac{1-x^4}{x^4}+\dfrac{1}{2}\cdot\dfrac{x^2}{\sqrt{1-x^4}}+ \dfrac{1}{2}\cdot\dfrac{x^2}{\sqrt{1-x^4}}\\ &\geqq 1+3\sqrt[3]{\dfrac{1-x^4}{x^4}\cdot\dfrac{1}{2}\cdot\dfrac{x^2}{\sqrt{1-x^4}}\cdot \dfrac{1}{2}\cdot\dfrac{x^2}{\sqrt{1-x^4}}}=1+\dfrac{3}{2^{2/3}}\\ &\left(\mathrm{equal\ sign}\Leftrightarrow \ \dfrac{1-x^4}{x^4} =\dfrac{1}{2}\cdot\dfrac{x^2}{\sqrt{1-x^4}}\Leftrightarrow\ x=\dfrac{2^{1/6}}{\sqrt[4]{1+2^{2/3} } }\in(0,1)\right)\\ &\mathrm{Hence,}\ \mathrm{the\ minimum\ value\ of\ the\ given\ expression\ is}\ \underline{ 1+\dfrac{3}{2^{2/3}}} \end{align}

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