961

已知正实数 ${\displaystyle a,b,c}$, 满足 ${\displaystyle abc=2}$, 若 ${\displaystyle max\{a,b,c\}\le 2}$, 求 ${\displaystyle a+b+c-{\displaystyle \frac{1}{a}-\frac{1}{b}-\frac{1}{c}}}$ 的最小值.

使用拉格朗日乘子法. 记 $$f(a,b,c;\lambda )={\displaystyle a+b+c-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}+\lambda (\ln a+\ln b+\ln c-\ln 2).}$$ 令 $$\begin{array}{r}\frac{\partial f}{\partial a}=\frac{a^2+\lambda a+1}{a^2}=0,\end{array}$$
同理$$\begin{array}{r}{b}^2+\lambda b+1=c^2+\lambda c+1=0.\end{array}$$
解得${\displaystyle (a,b,c)={\displaystyle (\sqrt[3]{2},\sqrt[3]{2},\sqrt[3]{2}),(2,2,\frac{1}{2})}}$(如果${\displaystyle a,b,c}$不全相等，则由韦达定理可知不相等的两者乘积为${\displaystyle 1}$.), 所求式的值分别为${\displaystyle {\displaystyle 3(\sqrt[3]{2}-\frac{1}{\sqrt[3]{2}})<\frac{3}{2}}}$.因此答案是${\displaystyle {\displaystyle 3(\sqrt[3]{2}-\frac{1}{\sqrt[3]{2}})}}$.