836

已知 ${\displaystyle a,b,c\in {\mathbb{R}}^{+},M>1,abc=M}$, 且 ${\displaystyle max(a,b,c)\le M}$, 求 ${\displaystyle N=a+b+c-{\displaystyle \frac{1}{a}-\frac{1}{b}-\frac{1}{c}}}$ 的最小值.

由条件得 ${\displaystyle M=abc\ge Mbc\Rightarrow bc\le 1}$. 此外 ${\displaystyle M=abc\le M^3\Rightarrow M\ge 1}$. 进而$$\begin{array}{rl}& N=a-\frac{1}{a}+(b+c)(1-\frac{1}{bc})\\ \ge & a-\frac{1}{a}+2\sqrt{bc}(1-\frac{1}{bc})=a-\frac{1}{a}+2\sqrt{\frac{M}{a}}(1-\frac{a}{M})\end{array}$$ 记 ${\displaystyle f(x)=x^2-{\displaystyle \frac{1}{x^2}+2(\frac{\sqrt{M}}{x}-\frac{x}{\sqrt{M}})}}$, ${\displaystyle 0\le x\le \sqrt{M}}$, 则 ${\displaystyle f^{\prime }(x)={\displaystyle \frac{(x^3-\sqrt{M})(x\sqrt{M}-1)}{x^3\sqrt{M}}}}$, 则 ${\displaystyle f(x)}$ 在 ${\displaystyle {\displaystyle (0,\frac{1}{\sqrt{M}})}}$, ${\displaystyle (\sqrt[6]{M},\sqrt{M})}$ 上增，在 ${\displaystyle {\displaystyle (\frac{1}{\sqrt{M}},\sqrt[6]{M})}}$ 上减，因此 $$\begin{array}{r}N=f(x^2)\ge f(\sqrt[6]{M})=3(\sqrt[3]{M}-\frac{1}{\sqrt[3]{M}}),\end{array}$$ 取等条件为 ${\displaystyle a=b=c=\sqrt[3]{M}}$.