01.4实数系的构造P044命题1.4.9
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设 $\displaystyle x\geq 0$, 则
\begin\{aligned\} 0<\alpha\leq 1&\Rightarrow (1+x)^\alpha\leq 1+\alpha x, \qquad(I)\\\\ \alpha\geq 1&\Rightarrow (1+x)^\alpha\geq 1+\alpha x,\qquad(II)\\\\ \alpha\geq 0&\Rightarrow (1+x)^\alpha\geq 1+\frac\{\alpha x\}\{1+x\}.\qquad(III) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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(1)、 对 $\displaystyle 0<\alpha\leq 1$, 取
\begin\{aligned\} \mathbb\{Z\}\ni m > \frac\{1\}\{1-\alpha\}, n=+1, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则
\begin\{aligned\} \alpha\leq \frac\{n\}\{m\}\leq \alpha+\frac\{1\}\{m\}<1\Leftrightarrow \frac\{n-1\}\{m\}\leq \alpha\leq \frac\{n\}\{m\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而
\begin\{aligned\} (1+x)^\alpha&\leq (1+x)^\frac\{n\}\{m\}=\left\^\frac\{1\}\{m\} \stackrel\{\tiny\mbox\{均值\}\}\{\leq\} \frac\{(m-n)+n(1+x)\}\{m\}\\\\ &=1+\frac\{n\}\{m\}x=1+\alpha x+\frac\{x\}\{m\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle m$ 的任意性即知 $\displaystyle (1+x)^\alpha\leq 1+\alpha x$.
(2)、
\begin\{aligned\} &(1+x)^\alpha\geq 1+\alpha x \Leftrightarrow 1+x\geq (1+\alpha x)^\frac\{1\}\{\alpha\} \Leftrightarrow (1+\alpha x)^\frac\{1\}\{\alpha\}\leq 1+\frac\{1\}\{\alpha\}\cdot \alpha x\Leftarrow (I). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、
\begin\{aligned\} (1+x)^\alpha\geq 1+\frac\{\alpha x\}\{1+x\} \Leftrightarrow& (1+x)^\alpha\geq \frac\{1+(1+\alpha)x\}\{1+x\}\\\\ \Leftrightarrow& (1+x)^\{1+\alpha\}\geq 1+(\alpha+1)x\Leftarrow (II). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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