正三棱锥的体积 $V = \dfrac{{\sqrt 2 }}{3}$，求正三棱锥的表面积的最小值；
【难度】
【出处】
2010年清华大学等五校合作自主选拔通用基础测试数学试题
【标注】
【答案】
$\sqrt 3 \cdot {2^{\frac{4}{3}}}$
【解析】
设正三棱锥的底面边长为 $a$，高为 $h$，则$$\dfrac{1}{3} \cdot \dfrac{{\sqrt 3 }}{4}{a^2} \cdot h = \dfrac{{\sqrt 2 }}{3},$$所以 $h = \dfrac{{4\sqrt 6 }}{{3{a^2}}}$，于是\[\begin{split}\qquad\qquad\qquad S &= \dfrac{{\sqrt 3 }}{4}{a^2} + 3 \cdot \dfrac{1}{2} \cdot a \cdot \sqrt {{h^2} + \dfrac{{{a^2}}}{{12}}} \\&= \dfrac{{\sqrt 3 }}{4}\left( {{a^2} + \sqrt {\dfrac{{128}}{{{a^2}}} + {a^4}} } \right)
\\&= \dfrac{{\sqrt 3 }}{4}\left( {{a^2} + \sqrt {\underbrace {\dfrac{{16}}{{{a^2}}} + \dfrac{{16}}{{{a^2}}} + \cdots + \dfrac{{16}}{{{a^2}}}}_8 + {a^4}} } \right) \\&\geqslant \dfrac{{\sqrt 3 }}{4}\left( {{a^2} + 3\cdot\dfrac{{{2^{\frac{{16}}{9}}}}}{{{a^{\frac{2}{3}}}}}} \right)\\&= \dfrac{{\sqrt 3 }}{4}\left( {{a^2} + \dfrac{{{2^{\frac{{16}}{9}}}}}{{{a^{\frac{2}{3}}}}} + \dfrac{{{2^{\frac{{16}}{9}}}}}{{{a^{\frac{2}{3}}}}} + \dfrac{{{2^{\frac{{16}}{9}}}}}{{{a^{\frac{2}{3}}}}}} \right) \\&\geqslant \dfrac{{\sqrt 3 }}{4} \cdot 4 \cdot {2^{\frac{4}{3}}}\\&= \sqrt 3 \cdot {2^{\frac{4}{3}}},\end{split}\]当且仅当 ${a^2} = {2^{\frac{4}{3}}}$ 时取得等号．
\\&= \dfrac{{\sqrt 3 }}{4}\left( {{a^2} + \sqrt {\underbrace {\dfrac{{16}}{{{a^2}}} + \dfrac{{16}}{{{a^2}}} + \cdots + \dfrac{{16}}{{{a^2}}}}_8 + {a^4}} } \right) \\&\geqslant \dfrac{{\sqrt 3 }}{4}\left( {{a^2} + 3\cdot\dfrac{{{2^{\frac{{16}}{9}}}}}{{{a^{\frac{2}{3}}}}}} \right)\\&= \dfrac{{\sqrt 3 }}{4}\left( {{a^2} + \dfrac{{{2^{\frac{{16}}{9}}}}}{{{a^{\frac{2}{3}}}}} + \dfrac{{{2^{\frac{{16}}{9}}}}}{{{a^{\frac{2}{3}}}}} + \dfrac{{{2^{\frac{{16}}{9}}}}}{{{a^{\frac{2}{3}}}}}} \right) \\&\geqslant \dfrac{{\sqrt 3 }}{4} \cdot 4 \cdot {2^{\frac{4}{3}}}\\&= \sqrt 3 \cdot {2^{\frac{4}{3}}},\end{split}\]当且仅当 ${a^2} = {2^{\frac{4}{3}}}$ 时取得等号．
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