📝 题目
例 7 设 $0 < x < 1$ . 求证: $x{\mathrm{e}}^{-x} > \frac{1}{x}{\mathrm{e}}^{-\frac{1}{x}}$ .
💡 答案与解析
证 先证原不等式两边取对数得到的不等式, 即证
$$ \ln x - x > - \ln x - \frac{1}{x}\;\left( {0 < x < 1}\right) $$
或
$$ 2\ln x - x + \frac{1}{x} > 0\;\left( {0 < x < 1}\right) . \tag{6.11} $$
为此令 $f\left( x\right) = 2\ln x - x + \frac{1}{x}$ ,则有 $f\left( 1\right) = 0$ ,且
$$ {f}^{\prime }\left( x\right) = \frac{2}{x} - 1 - \frac{1}{{x}^{2}} = - \frac{{\left( x - 1\right) }^{2}}{{x}^{2}} < 0\left( {0 < x < 1}\right) \Rightarrow f\left( x\right) \downarrow $$
$$ \Rightarrow f\left( x\right) > f\left( 1\right) = 0\left( {0 < x < 1}\right) \Rightarrow \left( {6.11}\right) \text{ 式. } $$
再利用函数 ${\mathrm{e}}^{t}$ 的严格递增性,由不等式 (6.11) 推出
$$ {x}^{2}{\mathrm{e}}^{-x} \cdot {\mathrm{e}}^{\frac{1}{x}} > 1\text{ ,即 }x{\mathrm{e}}^{-x} > \frac{1}{x}{\mathrm{e}}^{-\frac{1}{x}}\;\left( {0 < x < 1}\right) . $$