954

若不等式 ${\displaystyle {\displaystyle \ln x\le \frac{a}{x}+b\le {\mathrm{e}}^x(a,b\in \mathbb{R})}}$ 对任意的 ${\displaystyle x\in {\displaystyle [1,\frac{3}{2}]}}$ 恒成立，求 ${\displaystyle a}$ 的最小值.

首先代入边界条件$$\begin{array}{r}-\mathrm{e}\le -(a+b)\le 0,\ln \frac{3}{2}\le \frac{2a}{3}+b\le {\mathrm{e}}^{\frac{3}{2}},\end{array}$$
然后相加得$$\begin{array}{r}-\frac{a}{3}\le {\mathrm{e}}^{\frac{3}{2}}\Rightarrow a\ge -3{\mathrm{e}}^{\frac{3}{2}}.\end{array}$$
取等时 ${\displaystyle (a,b)={\displaystyle (-3{\mathrm{e}}^{\frac{3}{2}},3{\mathrm{e}}^{\frac{3}{2}})}}$.

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如果这是一个大题，下面要验证充分性：$$\begin{array}{r}\ln x\le 3{\mathrm{e}}^{\frac{3}{2}}(1-\frac{1}{x})\le {\mathrm{e}}^x,\mathrm{\forall }x\in [1,\frac{3}{2}].\end{array}$$

首先记 ${\displaystyle f(x)={\displaystyle \ln x-3{\mathrm{e}}^{\frac{3}{2}}(1-\frac{1}{x})}}$, 则 ${\displaystyle f^{\prime }(x)={\displaystyle \frac{1}{x}-3{\mathrm{e}}^{\frac{3}{2}}\frac{1}{x^2}}}$, 故$$\begin{array}{r}F(x)\le f\left(3{\mathrm{e}}^{\frac{3}{2}}\right)=\ln 3{\mathrm{e}}^{\frac{3}{2}}-(3{\mathrm{e}}^{\frac{3}{2}}-1)\le 0.\end{array}$$
再记 ${\displaystyle g(x)={\displaystyle 3{\mathrm{e}}^{\frac{3}{2}-x}(1-\frac{1}{x})}}$, 则 ${\displaystyle {\displaystyle g^{\prime }(x)=3{\mathrm{e}}^{\frac{3}{2}-x}\left(\frac{-x^2+x+1}{x^2}\right)>0,\mathrm{\forall }x\in [1,\frac{3}{2}]}}$, 故 ${\displaystyle g(x)\le {\displaystyle g\left(\frac{3}{2}\right)=1}}$.

这样充分性得证，${\displaystyle a}$ 最小值为 ${\displaystyle -3{\mathrm{e}}^{\frac{3}{2}}}$.