方企勤 第二章 一元函数微分学 第1题
📝 题目
例 1 求证: 当 $x > 0$ 时,不等式 $\ln \left( {1 + x}\right) < \frac{x}{\sqrt{1 + x}}$ 成立.
💡 答案解析
证法 1 作辅助函数 $f\left( x\right) = \ln \left( {1 + x}\right) - \frac{x}{\sqrt{1 + x}}$ ,则 $f\left( 0\right) = 0$ ,
$$ f\left( x\right) = \ln \left( {1 + x}\right) - \sqrt{1 + x} + \frac{1}{\sqrt{1 + x}}, $$
$$ {f}^{\prime }\left( x\right) = \frac{1}{1 + x} - \frac{1}{2\sqrt{x + 1}} - \frac{1}{2\sqrt{x + 1}\left( {x + 1}\right) } $$
$$ = \frac{2\sqrt{x + 1} - x - 2}{2\sqrt{x + 1}\left( {x + 1}\right) } = - \frac{x + 1 - 2\sqrt{x + 1} - 1}{2\sqrt{x + 1}\left( {x + 1}\right) } $$
$$ = - \frac{{\left( \sqrt{x + 1} - 1\right) }^{2}}{2\sqrt{x + 1}\left( {x + 1}\right) } < 0\;\left( {x > 0}\right) , \tag{6.1} $$
从而 $f\left( x\right) \downarrow \Rightarrow f\left( x\right) < f\left( 0\right) = 0$ . 即证得不等式.
证法 2 作辅助函数 $f\left( x\right) = \sqrt{1 + x}\ln \left( {1 + x}\right) - x$ ,则 $f\left( 0\right) = 0$ ,
$$ {f}^{\prime }\left( x\right) = \frac{\ln \left( {1 + x}\right) }{2\sqrt{x + 1}} - \frac{\sqrt{x + 1}}{x + 1} = \frac{\ln \left( {1 + x}\right) + 2 - 2\sqrt{x + 1}}{2\sqrt{x + 1}}. $$
(6.2)${f}^{\prime }\left( x\right)$ 的表达式 (6.2) 的右端分母肯定是正的,可单独考虑分子. 令 $g\left( x\right) = \ln \left( {1 + x}\right) + 2 - 2\sqrt{x + 1}$ ,则
$$ {g}^{\prime }\left( x\right) = \frac{1}{1 + x} - \frac{1}{\sqrt{1 + x}} = \frac{\sqrt{x + 1} - \left( {x + 1}\right) }{\left( {x + 1}\right) \sqrt{x + 1}} < 0\;\left( {x > 0}\right) . $$
由此可见, $g\left( x\right) \downarrow \Rightarrow g\left( x\right) < g\left( 0\right) = 0$ . 从而,由 (6.2) 式,有
$$ {f}^{\prime }\left( x\right) = \frac{g\left( x\right) }{2\sqrt{x + 1}} \leq 0. $$
于是由 $f\left( x\right) \downarrow$ 推出 $f\left( x\right) \leq f\left( 0\right) = 0$ . 即证得不等式.
证法 3 作变量代换 $t = \sqrt{1 + x}$ ,则
$$ \ln \left( {1 + x}\right) \leq \frac{x}{\sqrt{1 + x}}\left( {x > 0}\right) \Leftrightarrow 2\ln t \leq t - \frac{1}{t}\left( {t > 1}\right) . $$
令 $g\left( t\right) = t - \frac{1}{t} - 2\ln t$ ,则
$$ {g}^{\prime }\left( t\right) = 1 + \frac{1}{{t}^{2}} - \frac{2}{t} = \frac{{t}^{2} - {2t} + 1}{{t}^{2}} = {\left\lbrack \frac{t - 1}{t}\right\rbrack }^{2} > 0\;\left( {t > 1}\right) . $$
由此可见,当 $t > 1$ 时 $g\left( t\right)$ 为严格单调增加函数,又 $g\left( 1\right) = 0$ ,故有 $g\left( t\right) > g\left( 1\right) = 0$ . 即证得不等式.
证法 4 应用柯西中值定理. 对于 $\forall x > 0$ ,存在 $\xi \in \left( {0,x}\right)$ ,使得
$$ \frac{\ln \left( {1 + x}\right) }{\frac{x}{\sqrt{1 + x}}} = \frac{\frac{1}{1 + \xi }}{\frac{\sqrt{1 + \xi } - \frac{\xi }{2\sqrt{1 + \xi }}}{1 + \xi }} = \frac{\sqrt{1 + \xi }}{1 + \frac{\xi }{2}} $$
$$ = {\left\lbrack \frac{1 + \xi }{{\left( 1 + \frac{\xi }{2}\right) }^{2}}\right\rbrack }^{\frac{1}{2}} < 1. $$
即证得不等式.