数学人眼中的湖北

时间：2020-04-09 12:53:35 收藏：0 阅读：114

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\author{Yongtao Li}
\address{School of Mathematics and Statistics, Central South University,
Changsha, Hunan, 410083, P.R. China}
\email{ytli0921@csu.edu.cn}

\begin{document}

\begin{center}
{\LARGE 数学人眼中的湖北(待修改)}\\
曾熊, 李永涛
\end{center}

\section{2015年湖北卷理科}
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对于某些不等式, 尤其是左边是数列前$n$项的和式(积式), 右边是某个常数,
我们应用归纳法, 一般来说是没有办法利用归纳假设做递推的,
这个时候, 需要待证不等式的右边进行改造和加强. 下面以2015年湖北理科数学压轴题为例.

\begin{exam}[2015年湖北卷, 理科第22题] \label{2015hubeili}
已知数列$\{a_n\}$的各项均为正数, 且$b_n=n\left(1+\frac{1}{n}\right)^na_n(n\in \mathbb{N})$,
其中$e$为自然对数的底数.
\begin{itemize}
\item[(1)] 求函数$f(x)=1+x-e^x$的单调区间, 并比较$(1+\frac{1}{n})^n$与$e$的大小.

\item[(2)] 计算$\frac{b_1}{a_1},\frac{b_1b_2}{a_1a_2},\frac{b_1b_2b_3}{a_1a_2a_3}$, 由此推测计算
$\frac{b_1b_2\cdots b_n}{a_1a_2\cdots a_n}$的公式, 并给出证明.

\item[(3)] 令$c_n=(a_1a_2\cdots a_n)^{\frac{1}{n}}$, 数列$\{a_n\}$和$\{c_n\}$的前$n$项和分别记为
$S_n,T_n$, 试证明\sld ~$T_n<eS_n$.
\end{itemize}
\end{exam}
我们下面仅仅说明第(3)小问, 即需要证明
\begin{equation} \label{e137} \sum\limits_{k=1}^n\sqrt[k]{a_1a_2\cdots a_k}
\leqslant e\sum\limits_{k=1}^na_k, \end{equation}
上式称为Carleman不等式, 进一步, 我们还可以证明, 右端的常数$e$是最佳的,
也就是说, 没有更小的常数使得上式成立.

\begin{proof}[证明一]
令$b_k=\dfrac{(k+1)^k}{k^{k-1}}(k=1,2,\ldots, n)$, 根据AM-GM不等式可得
\begin{align*}\sum\limits_{k=1}^n\sqrt[k]{a_1a_2\cdots a_k}
&=\sum\limits_{k=1}^n\frac{\sqrt[k]{(a_1b_1)(a_2b_2)\cdots (a_kb_k)}}{k+1}
\leqslant \sum\limits_{k=1}^n\frac{1}{k(k+1)}\left( \sum\limits_{i=1}^na_ib_i \right) \\
&= \sum\limits_{i=1}^n a_ib_i\sum\limits_{j=i}^n\frac{1}{j(j+1)}
= \sum\limits_{i=1}^n a_ib_i \left( \frac{1}{i}-\frac{1}{n+1}\right).\end{align*}
注意到
\[ b_i \left( \frac{1}{i}-\frac{1}{n+1}\right)<\frac{b_i}{i}=\left( 1+\frac{1}{i}\right)^i< e.\]
因此, 不等式(\ref{e137})成立.

我们接下来说明$e$是最佳的, 令$a_k=1/k$, 于是
\[ \lim\limits_{n\to \infty}{\sum\limits_{k=1}^n\frac{1}{k}}\biggm/ {\sum\limits_{k=1}^n
\frac{1}{\sqrt[n]{n!}}}=\lim\limits_{n\to \infty}\frac{\sqrt[n]{n!}}{n}=e.\]
所以, $e$是使得不等式(\ref{e137})成立的最小常数.
\end{proof}

\begin{proof}[证明二]
证明一构造的数列$b_k$不是唯一的, 注意到
\[ \sqrt[k]{a_1a_2\cdots a_k}=\frac{1}{\sqrt[k]{k!}}\cdot \sqrt[k]{(1\cdot a_1)(2\cdot a_2)\cdots
(k\cdot a_k)}\leqslant \frac{1}{\sqrt[k]{k!}}\cdot \frac{a_1+2a_2+\cdots +ka_k}{k}.\]
结合不等式$\sqrt[n]{n!}>(n+1)/e$(见数学归纳法章节), 于是
\[ \sum\limits_{k=1}^n\sqrt[k]{a_1a_2\cdots a_k}<e\cdot \sum\limits_{k=1}^n
\frac{a_1+2a_2+\cdots +ka_k}{k(k+1)}=e\cdot \sum\limits_{k=1}^n \left(
\sum\limits_{j=k}^n\frac{k}{j(j+1)}\right)a_k<e\sum\limits_{k=1}^na_k.\qedhere \]
\end{proof}

\begin{proof}[证明三]
根据Hardy-Landau不等式(不懂就百度吧), 我们有
\[ \sum\limits_{k=1}^n\left( \frac{a_1^{1/p}+a_2^{1/p}+\cdots +a_k^{1/p}}{k}\right)^p
\leqslant \left( \frac{p}{p-1}\right)^p\sum\limits_{k=1}^na_k.\]
我们令$p\to +\infty$得
\[\lim\limits_{p\to +\infty}\left( \frac{a_1^{1/p}+a_2^{1/p}+\cdots +a_k^{1/p}}{k}\right)^p
=\sqrt[k]{a_1a_2\cdots a_k},\]
以及
\[ \lim\limits_{p\to +\infty}\left( \frac{p}{p-1}\right)^p=e. \]
所以, 不等式(\ref{e137})成立.
\end{proof}

\subparagraph{注}
对于AM-GM不等式, 我们在利用它放缩时, 通常会对它加以调整, 通过引入待定的参数使得不等式更加精确.
比如, 当$\lambda_k>0$时,
\[ \sqrt[n]{a_1a_2\cdots a_n}=\frac{\sqrt[n]{(\lambda_1a_1)(\lambda_2a_2)
\cdots (\lambda_na_n)}}{\sqrt[n]{\lambda_1\lambda_2\cdots \lambda_n}}
\leqslant \frac{\frac{1}{n}\sum\limits_{k=1}^n\lambda_ka_k}{\sqrt[n]{
\lambda_1\lambda_2\cdots \lambda_n}}.\]
特别地, 取$\lambda_k=k$时有
\[ \sqrt[n]{a_1a_2\cdots a_n}=\frac{\sqrt[n]{a_1(2a_2)\cdots (na_n)}}{\sqrt[n]{n!}}
\leqslant \frac{1}{\sqrt[n]{n!}}\cdot \frac{1}{n}\sum\limits_{k=1}^nka_k.\]

对于第(3)小问, 其难度是相当大的, 远远超出了中学生的能力范围, 属于竞赛类选手的难度.
直接证明不等式(\ref{e137})比较不容易, 但是我们对其右端改造后, 就可以很轻松地利用数学归纳法,
对, 确实是很轻松.

\begin{prop}
设$a_1,a_2,\ldots ,a_n$为非负实数, $e$为自然对数的底数, 则
\begin{equation}\label{e138} \sum\limits_{k=1}^n\sqrt[k]{a_1a_2\cdots a_k}
\leqslant e\sum\limits_{k=1}^na_k - n\sqrt[n]{a_1a_2\cdots a_n}.\end{equation}
\end{prop}

\begin{proof}[证明]
利用归纳法就等价于证明
\[ ea_n+(n-1)\sqrt[n-1]{a_1a_2\cdots a_{n-1}}\geqslant (n+1)\sqrt[n]{a_1a_2\cdots a_n}.\]
由AM-GM不等式可得
\[ ea_n+(n-1)\sqrt[n-1]{a_1a_2\cdots a_{n-1}}\geqslant n\sqrt[n]{ea_1a_2\cdots a_n}.\]
再结合不等式$e>\left( 1+\frac{1}{n}\right)^n$即可.
\end{proof}

当然, 这样的例子还有很多, 数学归纳法起着很巧妙绝伦的作用.

\begin{prop}[羊明亮]
设$x_1,x_2,\ldots ,x_n$为任意实数, 证明\sld
\[ \sum\limits_{k=1}^n\left( \frac{1}{k}\sum\limits_{j=1}^kx_j\right)^2\leqslant
\sum\limits_{k=1}^n(k+1)x_k^2.\]
\end{prop}

\subparagraph{注}
同样地, 利用归纳法可以证明如下不等式.
\[ \sum\limits_{k=1}^n\left( \frac{1}{k}\sum\limits_{j=1}^kx_j\right)^2 \leqslant \sum\limits_{k=1}^n(k+1)x_k^2 - \frac{1}{n}\left(\sum\limits_{k=1}^nx_k \right)^2.\]

\begin{prop}[2005年国家队选拔赛试题]
设$a_1,a_2,\ldots ,a_n$为正实数, 试证明\sld
\[ \left( \frac{\sum\limits_{j=1}^n\sqrt[j]{a_1a_2\cdots a_j}}{\sum\limits_{j=1}^na_j}\right)^{\!\!1/n}
+\frac{\sqrt[n]{a_1a_2\cdots a_n}}{\sum\limits_{j=1}^n\sqrt[j]{a_1a_2\cdots a_j}}\leqslant
\frac{n+1}{n}.\]
\end{prop}

\noindent
{\bf 注}~
该不等式是Carleman不等式的加强(为什么).

下面几例是类似的交换求和顺序的技巧.

\begin{prop}
给出最佳常数$C$, 使得对任意正数$a_1,a_2,\ldots,a_n$, 下述不等式成立.
\[ \sum\limits_{k=1}^n\frac{k}{\sum\limits_{j=1}^k\frac{1}{a_j}}\leqslant C\sum\limits_{k=1}^na_k.\]
\end{prop}

\begin{proof}[证明]
根据Cauchy-Schwarz不等式可得
\[ \left( \sum\limits_{j=1}^k\frac{1}{a_j}\right)\left( \sum\limits_{j=1}^kj^2a_j\right)\geqslant
\left( \sum\limits_{j=1}^kj\right)^2=\left[\frac{k(k+1)}{2}\right]^2.\]
整理便有
\[ \frac{k}{\sum\limits_{j=1}^k\frac{1}{a_j}}\leqslant \frac{4}{k(k+1)^2}\left( \sum\limits_{j=1}^kj^2a_j\right).\]
所以
\begin{align*}
\sum\limits_{k=1}^n\frac{k}{\sum\limits_{j=1}^k\frac{1}{a_j}}
&\leqslant \sum\limits_{k=1}^n\frac{4}{k(k+1)^2}\left( \sum\limits_{j=1}^kj^2a_j\right) \\
&=2\sum\limits_{j=1}^nj^2a_j\sum\limits_{k=j}^n\frac{2}{k(k+1)^2}\\
&<2\sum\limits_{j=1}^nj^2a_j\sum\limits_{k=j}^n\left( \frac{1}{k^2}-\frac{1}{(k+1)^2}\right) \\
&=2\sum\limits_{j=1}^nj^2a_j\left[ \frac{1}{j^2}-\frac{1}{(n+1)^2}\right] \\
&<2\sum\limits_{j=1}^na_j.\end{align*}
因此, 可取$C=2$, 下面说明$2$是最佳系数. 令$a_j=1/j(j=1,2,\cdots,n)$,
\[ \sum\limits_{k=1}^n\frac{k}{\sum\limits_{j=1}^k\frac{1}{a_j}}=\sum\limits_{k=1}^n
\frac{k}{\frac{1}{2}k(k+1)}=2\sum\limits_{k=1}^n\frac{1}{k+1}\leqslant C\sum\limits_{k=1}^n\frac{1}{k}.\]
由极限理论可知, 常数$C$不小于$2$.
\end{proof}

令$a_i=1/x_i$, 命题即为2005年美国数学月刊上第11145号征解问题:
\begin{exam}[AMM, 11145] \label{exam1321}
设$x_1,x_2,\ldots ,x_n$均为正数, 试证明\sld
\[ \sum\limits_{k=1}^n\frac{k}{x_1+x_2+\cdots +x_k}
\leqslant 2\sum\limits_{k=1}^n\frac{1}{x_k}. \]
\end{exam}

\noindent
{\bf 注}~
从该不等式可以看出\\
(1) 若$x_k>0$, 且$\sum\limits_{n=1}^{\infty}\frac{1}{x_n}$收敛,
则级数
$ \sum\limits_{n=1}^{\infty}\frac{n}{x_1+x_2+\cdots +x_n}$
也是收敛的. \\
下面这个不等式留给读者.
\[ \sum\limits_{k=1}^n\frac{2k+1}{x_1+x_2+\cdots +x_k}
\leqslant 4\sum\limits_{k=1}^n\frac{1}{x_k}.\]

\begin{prop}
设$a_k\geqslant 0(k=1,2,\ldots ,n)$, 则对每个正整数$m$有
\[ \sum\limits_{k=1}^n\sqrt[k]{a_1a_2\cdots a_k}\leqslant \frac{1}{m}\sum\limits_{k=1}^n
a_k\left( \frac{k+m}{k}\right)^k.\]
\end{prop}

\section{2014年湖北卷理科}

\begin{exam}[2014湖北卷, 理科第22题] \label{2014hubeili}
\quad \\
设$\pi $为圆周率, $e=2.71828\cdots $为自然对数的底数.
\begin{itemize}
\item[(1)] 求函数$f(x)=\frac{\ln x}{x}$的单调区间.

\item[(2)] 求$e^3,3^e,e^{\pi},\pi^e,3^{\pi}, \pi^3$这六个数的最大数与最小数.

\item[(3)] 将$e^3,3^e,e^{\pi},\pi^e,3^{\pi}, \pi^3$这六个数按从小到大的顺序排列, 并证明你的结论.
\end{itemize}
\end{exam}

\begin{proof}[证明]
(3) 我们只需要比较$e^3$与$\pi^e$, $e^{\pi}$与$\pi^3$的大小,
这等价于比较$3$与$e\ln \pi $, $\pi $与$3\ln \pi$的大小.
我们下面寻求对$\ln \pi$的估计, 根据(1)可知, 函数$f(x)=\frac{\ln x}{x}$在区间$(0,e)$上单调递增,
在区间$(e,+\infty )$上单调递减. 注意到$\frac{e^2}{\pi }<e$, 于是
\[ \frac{\ln \frac{e^2}{\pi }}{\frac{e^2}{\pi}}<\frac{\ln e}{e}\Rightarrow
\ln \pi >2-\frac{e}{\pi }. \]
于是$e\ln \pi >e(2-\frac{e}{\pi })>2.7\times (2-0.88)=3.024>3$, 即$\pi^e >e^3$. 另一方面,
$3\ln \pi >3(2-\frac{e}{\pi })>3\times (2-0.88)=3.36 >\pi $, 即$\pi^3 >e^{\pi}$.
\end{proof}

我们下面得到关于$\ln \pi$的上界.
\[ \frac{\ln \frac{\pi^2}{e}}{\frac{\pi^2}{e}}>\frac{\ln e}{e} \Rightarrow
\ln \pi <\frac{\pi^2}{2e^2}+\frac{1}{2}\approx 1.17. \]
关于常数$e$与$\pi$还有很多有趣的知识, 例如$e^{\pi }$被称为盖尔范德常数, 已经被证明是超越数,
奇怪的是, $\pi^e$却了解甚少, 目前还没有被证明是否是无理数.

\section{2012年湖北卷理科}

\begin{exam}[2012年湖北卷, 理科第21题] \label{2012hubeili}\quad \\
(1) 设$\!f(x)\!=\!rx\!-\!x^r\!+(1\!-\!r),x>0$, 其中$\!r\!$为有理数,
且$\!0<r<1$. 求$\!f(x)\!$的最小值.\\
(2) 试用(1) 的结果证明如下命题\sld 设$\!a_1,a_2\!\geqslant\! 0$, $b_1,b_2\!$为正有理数,
且$\!b_1\!+\!b_2=1$, 则
\[ a_1^{b_1}a_2^{b_2}\leqslant a_1b_1+a_2b_2.\]
(3) 请将(2) 中的命题推广到一般形式, 利用数学归纳法证明你所推广的命题.
\end{exam}

\section{2011年湖北卷理科}

\begin{exam}[2011年湖北卷, 理科第21题] \label{2011hubeili}\quad \\
(1) 已知函数$f(x)=\ln x-x+1,x\in (0,+\infty)$, 求函数$f(x)$的最大值.\\
(2) 设$a_k,b_k$均为正实数, 求证\sld \\
$\mathrm{(i)}$ 若$a_1b_1+a_2b_2+\cdots+a_nb_n\leqslant b_1+b_2+\cdots +b_n$, 则
\[ a_1^{b_1}a_2^{b_2}\cdots a_n^{b_n}\leqslant 1.\]
$\mathrm{(ii)}$ 若$b_1+b_2+\cdots +b_n=1$, 则
\[ \frac{1}{n}\leqslant b_1^{b_1}b_2^{b_2}\cdots b_n^{b_n}\leqslant b_1^2+b_2^2+\cdots +b_n^2.\]
\end{exam}

\begin{proof}[证明]
(2) 根据Jensen不等式有
\[\frac{\sum\limits_{k=1}^nb_k\cdot \ln a_k}{\sum\limits_{k=1}^nb_k}\leqslant \ln \left(\frac{\sum\limits_{k=1}^nb_k\cdot a_k}{\sum\limits_{k=1}^nb_k}\right)\leqslant \ln 1=0.\qedhere\]
\end{proof}

\section{2010年湖北卷理科}

首先, 引入一些预备知识.

\begin{thm} \label{thmthm}
对于任意的$x\geqslant -1$, 下述不等式成立,
\begin{equation}\label{eq1}
\frac{x}{1+x}\leqslant \ln (1+x) \leqslant x.\end{equation}
当且仅当$x=0$时等号成立. 作一个倒代换$x=1/y$可得如下常用不等式,
\begin{equation}\label{eq2}
\frac{1}{1+y}< \ln \left(1+\frac{1}{y} \right)<\frac{1}{y} ,
\text{~其中$y>0$或$y\leqslant -1$}.\end{equation}
\end{thm}

\begin{proof}[证明]
直接移项构造函数
\[ f(x)=\ln (x+1)-\frac{x}{1+x},\quad g(x)=x-\ln (x+1). \]
然后求导数, 判断函数的单调性, 接着利用单调性确定最值点.
\end{proof}

作为式(\ref{eq1})的一个直接应用是证明如下常见不等式:
\begin{gather} a<\frac{a-b}{\ln a-\ln b}<b,\quad 0<a<b.
\end{gather}
对于不等式(\ref{eq2}), 变形即有,
\begin{gather}
\frac{1}{1+y}<\ln (y+1)-\ln y<\frac{1}{y}<\ln y-\ln (y-1). \end{gather}
在上式中, 令$y=1,2,3,\ldots ,n$可得
\begin{equation} \label{eq2.5}
\left\{ \begin{aligned}
\ln 2-\ln 1<&1, \\
\ln 3-\ln 2<&\frac{1}{2} <\ln 2-\ln 1, \\
\ln 4-\ln 3<&\frac{1}{3} <\ln 3-\ln 2,\\
\cdots &\cdots\cdots \\
\ln (n+1) -\ln n <&\frac{1}{n}<\ln n -\ln (n-1).
\end{aligned}
\right.
\end{equation}
将上面所有不等式累加可得
\begin{equation}
\label{eq24} \ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}< 1+\ln n.
\end{equation}
同理可以推出下面两式(留给读者).
\begin{gather}
\frac{\ln 2}{2}\cdot \frac{\ln 3}{3}\cdot\frac{\ln 4}{4}\cdot \cdots \cdot \frac{\ln n}{n}<\frac{1}{n}.\\
\frac{1}{\ln 2}+\frac{1}{\ln 3}+\frac{1}{\ln 4}+\cdots +\frac{1}{\ln n}>\frac{3}{2}.\end{gather}
不等式(\ref{eq24})的几何直观如下.
\begin{center}
\begin{tikzpicture}[domain=-2:4,scale=1.2]
\draw[elegant,color=black,domain=0.3:4.8] plot (\x,{1/((\x))})
node[above ] {$\displaystyle \frac{1}{x}$};

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\fill[yellow!60!white] (1,0) -- (1.5,0) -- (1.5,0.66666) -- (1,0.66666) -- cycle;
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\fill[yellow!60!white] (3,0) -- (3.5,0) -- (3.5,0.29) -- (3,0.29) -- cycle;
\fill[yellow!60!white] (3.5,0) -- (4,0) -- (4,0.23) -- (3.5,0.23) -- cycle;

\draw[dashed] (0.5,0) -- (0.5,2) -- (0,2) ;
\draw[dashed] (1,0) -- (1,1) -- (0.5,1);
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\draw[dashed] (2,0) -- (2,0.5) -- (1.5,0.5) ;
\draw[dashed] (3.5,0) -- (3.5,0.29) -- (3,0.29)--(3,0) ;
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\draw (0,0) node[below left] {$O$};

\draw (0.5,0) node[below] {$1$};
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\draw (1.5,0) node[below] {$3$};
\draw (2,0) node[below] {$4$};
\draw (3.5,0) node[below] {$n\!\!-\!\!1$};
\draw (4,0) node[below=2pt] {$n$};

\shade[ball color=black](2.35,0.2) circle(0.5pt);
\shade[ball color=black](2.5,0.2) circle(0.5pt);
\shade[ball color=black](2.65,0.2) circle(0.5pt);

\draw[->] (-0.4,0) -- (5,0) node[below=2pt] {$x$};
\draw[->] (0,-0.4) -- (0,3.5) node[left] {$y$};
\end{tikzpicture}
\begin{tikzpicture}[domain=-2:4,scale=1.2]
\fill[yellow!60!white] (0.5,0) -- (1,0) -- (1,2) -- (0.5,2) -- cycle;
\fill[yellow!60!white] (1,0) -- (1.5,0) -- (1.5,1) -- (1,1) -- cycle;
\fill[yellow!60!white] (1.5,0) -- (2,0) -- (2,0.66666) -- (1.5,0.66666) -- cycle;
\fill[yellow!60!white] (2,0) -- (2.5,0) -- (2.5,0.5) -- (2,0.5) -- cycle;
\fill[yellow!60!white] (3.5,0) -- (4,0) -- (4,0.32) -- (3.5,0.32) -- cycle;
\fill[yellow!60!white] (4,0) -- (4.5,0) -- (4.5,0.27) -- (4,0.27) -- cycle;

\draw[dashed] (0.5,0) -- (0.5,2) -- (1,2)--(1,1) ;
\draw[dashed] (1,0) -- (1,1) -- (1.5,1) -- (1.5,0.66666);
\draw[dashed] (1.5,0) -- (1.5,0.66666) -- (2,0.66666)--(2,0.5) ;
\draw[dashed] (2,0) -- (2,0.5) -- (2.5,0.5) --(2.5, 0) ;

\draw[dashed] (3.5,0) -- (3.5,0.32) -- (4,0.32)--(4,0.27) ;
\draw[dashed] (4,0) -- (4,0.27) -- (4.5,0.27) -- (4.5,0);

\draw (0,0) node[below left] {$O$};

\shade[ball color=black](2.85,0.2) circle(0.5pt);
\shade[ball color=black](3,0.2) circle(0.5pt);
\shade[ball color=black](3.15,0.2) circle(0.5pt);

\draw[->] (-0.4,0) -- (5,0) node[below=2pt] {$x$};
\draw[->] (0,-0.4) -- (0,3.5) node[left] {$y$};
\draw[elegant,color=black,domain=0.3:4.8] plot (\x,{1/((\x))})
node[above ] {$\displaystyle \frac{1}{x}$};
\end{tikzpicture}
\end{center}

下面是定理\ref{thmthm}的加强.
\begin{thm}
(1) 当$x\in (0,+\infty)$时, 有$\displaystyle \frac{x}{1+\frac{1}{2}x}<\ln (1+x)<\frac{x}{\sqrt{1+x}}$.\\
(2) 当$x\in (-1,0)$时, 有$\displaystyle\frac{x}{\sqrt{1+x}}<\ln (1+x)<\frac{x}{1+\frac{1}{2}x}$.
\end{thm}

\begin{proof}[证明]
构造函数
\[ f(x)=\frac{x}{\sqrt{1+x}}-\ln (1+x),\quad x\in (-1,\infty ). \]
\[ g(x)=\ln (1+x)-\frac{x}{1+\frac{1}{2}x},\quad x\in (-1,+\infty ). \]
接下来留给读者.
\end{proof}

该定理的几何直观如下:\\
\begin{center}
\begin{tikzpicture}[domain=-2:4,scale=2]
\draw[->] (-1.5,0) -- (4,0) node[below] {$x$};
\draw[->] (0,-2.5) -- (0,2) node[below left] {$y$};
%\draw[very thin,color=gray] (-3,-3) grid (3,3);

\foreach \x in {-0.5,0.5,1.5,2.5,3.5}
\draw (\x,-0.5pt) -- (\x,0.5pt) ;
\foreach \x in {1,2,3}
\draw (\x,-1pt) -- (\x,1pt) node[below=4pt] {$\x$};

\foreach \y in {-1.5,-0.5,0.5,1.5}
\draw (-0.5pt,\y) -- (0.5pt,\y);
\foreach \y in {-2,-1,1}
\draw (-1pt,\y) -- (1pt,\y) node[left=3pt] {$\y$};

\draw[elegant,color=black,domain=-0.9:3] plot (\x,{(\x)/(sqrt((\x)+1))})
node[above] {$f(x)=\frac{x}{\sqrt{x+1}}$};

\draw[elegant,thick,color=red,domain=-0.9:3] plot(\x,{ln((\x)+1)}) node[right] {$g(x)=\ln (x+1)$};

\draw[elegant,dashed,color=black,domain=-0.9:3] plot (\x,{(\x)/((1/2)*(\x)+1)})
node[below] {$h(x)=\frac{x}{1+\frac{1}{2}x}$};

\draw[dashed,color=gray] (-1,-3) -- (-1,1.5);
\draw (-1,0) node[below left] {$-1$};
\shade[ball color=black](-1,0)circle(0.8pt);
\end{tikzpicture}
\end{center}

当$x={1}/{y},y>0$时, 有不等式
\begin{equation}\label{eqgen}
\frac{1}{y+\frac{1}{2}}<\ln \left( 1+\frac{1}{y}\right)=\ln (y+1)- \ln y<\frac{1}{\sqrt{y(y+1)}}
<\frac{1}{2}\left( \frac{1}{y}+\frac{1}{y+1}\right).
\end{equation}
上式的几何解释如下面左图.
\begin{center}
\begin{tikzpicture}[domain=-2:4,scale=1.4]
\fill[yellow!60!white] (0.7,0) -- (2,0) -- (2,0.5) -- (0.7,1.42857) -- cycle;

\draw[dashed] (2,0) -- (2,0.5) -- (0.7,1.42857) -- (0.7,0) ;

\draw (0,0) node[below left] {$O$};

\draw[dashed] (1.35,0) -- (1.35,0.740) ;

\draw[dashed] (0.7, 1.06) -- (2,0.39) ;

\draw (0.7,0) node[below] {$y$};
\draw (1.35,0) node[below] {$y\!+\!\frac{1}{2}$};
\draw (2,0) node[below] {$y\!+\!1$};

\draw[->] (-0.3,0) -- (3,0) node[below=2pt] {$x$};
\draw[->] (0,-0.3) -- (0,3) node[left] {$y$};

\draw[color=black,domain=0.4:2.5] plot (\x,{1/((\x))})
node[above ] {$\displaystyle \frac{1}{x}$};
\end{tikzpicture}
\begin{tikzpicture}[domain=-2:4,scale=1.4]
\fill[yellow!60!white] (0.5,2) -- (0.6,1.6666) -- (0.7,1.42857) -- (0.8,1.25)
-- (0.9,1.1111) -- (1,1) -- (1,2) -- cycle;
\fill[yellow!60!white] (1,1) -- (1.1,0.9090) -- (1.2,0.8333) -- (1.3,0.76923)--
(1.4,0.71428 ) -- (1.5,0.6666) -- (1.5,1) -- cycle;
\fill[yellow!60!white] (1.5,0.6666) -- (1.6,0.625) -- (1.7,0.58823) --
(1.8,0.5555) -- (1.9,0.52631) -- (2,0.5) -- (2,0.6666) -- cycle;
\fill[yellow!60!white] (2,0.5) -- (2.2,0.4545) -- (2.5,0.4) -- (2.5,0.5) -- cycle;
\fill[yellow!60!white] (3.5,0.32) -- (3.7,0.27) -- (4,0.25) -- (4,0.32) -- cycle;
\fill[yellow!60!white] (4,0.27) -- (4.5,0.2222) -- (4.5,0.27) -- cycle;

\draw[dashed] (3.5,0) -- (3.5,0.32) -- (4,0.32)--(4,0.27) ;
\draw[dashed] (4,0) -- (4,0.27) -- (4.5,0.27) -- (4.5,0);

\draw (0,0) node[below left] {$O$};

\shade[ball color=black](2.85,0.2) circle(0.5pt);
\shade[ball color=black](3,0.2) circle(0.5pt);
\shade[ball color=black](3.15,0.2) circle(0.5pt);

\draw[->] (-0.3,0) -- (5,0) node[below=2pt] {$x$};
\draw[->] (0,-0.5) -- (0,3) node[left] {$y$};

\draw[elegant,color=black,domain=0.35:4.8] plot (\x,{1/((\x))})
node[above ] {$\displaystyle \frac{1}{x}$};
\end{tikzpicture}
\end{center}

\begin{thm}[M. Shao \cite{MYS}]
对每个正整数$n\in \mathbb{N}^*$, 令
\[ y_n:=1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}- \ln (n+1).\]
则$\lim\limits_{n\to \infty} y_n =\gamma$~\kai{欧拉常数}\,, 且$1/2<\gamma <2-2\ln 2=0.61370\cdots $.
\end{thm}

\begin{proof}[证明]
首先, 很显然有$y_1<y_2<\cdots <y_n<\cdots$, 且
\[ y_n<1+\ln n-\ln (n+1)<1. \]
所以$y_n$严格单调递增(见上面右图), 且有上界, 于是极限存在. 又易知
\[ \lim\limits_{n\to \infty}
\left[ 1\!+\!\frac{1}{2}\!+\!\frac{1}{3}\!+\!\cdots \!+\!\frac{1}{n}\!-\! \ln (n\!+\!1) \right]
\!\!=\!\lim\limits_{n\to \infty }
\left[ 1\!+\!\frac{1}{2}\!+\!\frac{1}{3}\!+\!\cdots \!+\!\frac{1}{n}\!-\! \ln n \right]
\!\!:=\gamma .\]
注意到, 对每个正整数$k$有
\[ \frac{2}{2k+1}<\ln (k+1)-\ln k <\frac{1}{2}\left( \frac{1}{k}+\frac{1}{k+1}\right). \]
于是
\[ \frac{1}{2}\left( \frac{1}{k} -\frac{1}{k+1}\right)
<\frac{1}{k}-\Bigl[ \ln (k+1)-\ln k\Bigr] <
2\left( \frac{1}{2k}-\frac{1}{2k+1}\right). \]
将上式$k=1,2,\ldots ,n$求和可得
\[ \frac{1}{2}\left( 1 -\frac{1}{n+1}\right)<
y_n < 2 \left( \frac{1}{2}-\frac{1}{3} +\frac{1}{4}-\frac{1}{5} + \cdots +
\frac{1}{2n}-\frac{1}{2n+1}\right). \]
在上式中令$n\to \infty $, 且注意到
\[ \ln 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots . \qedhere \]
\end{proof}

利用上述不等式可以直接得到以下两个不等式.
\[ \frac{1}{1+\frac{1}{2}}+\frac{1}{2+\frac{1}{2}}+\frac{1}{3+\frac{1}{2}}+\cdots +
\frac{1}{n+\frac{1}{2}} <\ln (n+1). \]
\[ \frac{1}{\sqrt{1\cdot 2}}+\frac{1}{\sqrt{2\cdot 3}}+\frac{1}{\sqrt{3\cdot 4}}+\cdots +
\frac{1}{\sqrt{n\cdot (n+1)}}>\ln (n+1). \]

我们可以估计一些特殊的自然对数的值, 比如令$x=1$得
\begin{equation}\label{eq257}
\frac{2}{3}<\ln 2<\frac{\sqrt{2}}{2}. \end{equation}

\begin{exam}[2010湖北卷, 理科第21题] \label{2010hubeili}
已知函数$f(x)=ax+\frac{b}{x}+c(a>0)$的图象
在点$(1,f(x))$处的切线方程为$y=x-1$.\\
\yi 用$a$表示出$b,c$.\\
\er 若$f(x)\geqslant \ln x$在$[1,+\infty)$上恒成立, 求$a$的取值范围.\\
\san 证明\sld 对任意$n\ge 1$有$1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}>\ln (n+1)+\frac{n}{2(n+1)}$.
\end{exam}

湖北卷第(3)问的几何解释.
\begin{center}
\begin{tikzpicture}[domain=-2:4,scale=1.2]
\fill[yellow!60!white] (0.5,0) -- (1,0) -- (1,2) -- (0.5,2) -- cycle;
\fill[yellow!60!white] (1,0) -- (1.5,0) -- (1.5,1) -- (1,1) -- cycle;
\fill[yellow!60!white] (1.5,0) -- (2,0) -- (2,0.66666) -- (1.5,0.66666) -- cycle;
\fill[yellow!60!white] (2,0) -- (2.5,0) -- (2.5,0.5) -- (2,0.5) -- cycle;
\fill[yellow!60!white] (3.5,0) -- (4,0) -- (4,0.32) -- (3.5,0.32) -- cycle;
\fill[yellow!60!white] (4,0) -- (4.5,0) -- (4.5,0.27) -- (4,0.27) -- cycle;

\draw[dashed] (0.5,2) -- (1,1) -- (1.5,0.66666) -- (2,0.5) -- (2.5,0.4);

\draw[dashed] (3.5,0) -- (3.5,0.32) -- (4,0.32)--(4,0.27) ;
\draw[dashed] (4,0) -- (4,0.27) -- (4.5,0.27) -- (4.5,0);

\draw (0,0) node[below left] {$O$};

\shade[ball color=black](2.85,0.2) circle(0.5pt);
\shade[ball color=black](3,0.2) circle(0.5pt);
\shade[ball color=black](3.15,0.2) circle(0.5pt);

\begin{thm}
对任意的$x>-1$, 试证明\sld~ $\displaystyle x-\frac{1}{2}x^2\leqslant \ln (1+x)\leqslant x$.
\end{thm}

类似的问题还有许多, 下面简单列举几例.

\begin{exam}[2012年天津卷] \label{2012tianjinli}
设$f(x)=x-\ln (x+a)$的最小值为$0$, 其中$a>0$.\\
\yi 求$a$的值.\\
\er 若对任意的$x\in [0,+\infty )$, 有$f(x)\leqslant kx^2$成立, 求实数$k$的最小值. \\
\san 证明\sld~ $\displaystyle \sum\limits_{k=1}^n\frac{2}{2k-1}-\ln (2n+1)<2$.
\end{exam}

\begin{proof}[证明一]
(2) 令$g(x)=f(x)-kx^2=x-\ln (x+1)-kx^2$, 于是
\[ f‘(x)=\frac{-x[2kx-(1-2k)]}{x+1}, \quad x\geqslant 0. \]
令$g‘(x)=0$, 得到$x_1=0,x_2=\frac{1-2k}{2k}>-1$. \\
(i) 注意到$0\leqslant x-\ln (x+1)\leqslant kx^2$, 于是$k\leqslant 0$不符合题意. \\
(ii) 当$k\geqslant 1/2$时, $x_2\leqslant 0$, 于是在$[0,+\infty )$上恒有$g‘(x)\leqslant 0$. 因此,
对任意的$x\in [0,+\infty )$时, 有$g(x)\leqslant g(0)=0$, 即$f(x)\leqslant kx^2$在$[0,+\infty )$上恒成立,
故$k\geqslant 1/2$符合题意. \\
(iii) 当$0<k<1/2$时, 有$\frac{1-2k}{2k}>0$, 对于$x\in (0,\frac{1-2k}{2k} )$时, $g‘(x)\geqslant 0$,
$g(x)$在$(0,\frac{1-2k}{2k} )$上单调递增, 于是当$x\in (0,\frac{1-2k}{2k} )$时, $g(x)>g(0)=0$, 即
$f(x)>kx^2$, 因此$0<k<1/2$不符合题意. \\
(3) 首先, 当$n=1$时, 不等式即为$2-\ln 3<2$, 不等式成立. 当$n\geqslant 2$时, 注意到
\[ \ln (2n+1)=\sum\limits_{k=1}^n\ln \frac{2k+1}{2k-1}=\sum\limits_{k=1}^n \ln
\left(1+\frac{2}{2k-1}\right). \]
于是,
\[ \sum\limits_{k=1}^n\frac{2}{2k-1}-\ln (2n+1)=\sum\limits_{k=1}^n
\left[ \frac{2}{2k-1}-\ln \left(1+\frac{2}{2k-1}\right) \right]. \]
在第(2)小问中, 我们有$x-\ln (1+x)\leqslant kx^2(k\geqslant \frac{1}{2})$, 于是取$k$最小有
\begin{equation} x-\ln (1+x) \leqslant \frac{1}{2}x^2. \end{equation}
从而, 当$k\geqslant 2$时, 有
\[ \frac{2}{2k-1}-\ln \left(1+\frac{2}{2k-1}\right)\leqslant \frac{1}{2}\left(\frac{2}{2k-1}\right)^2
=\frac{2}{(2k-1)^2}<\frac{2}{(2k-3)(2k-1)}. \]
因此
\begin{align*}
\sum\limits_{k=1}^n\frac{2}{2k-1}-\ln (2n+1)
&<2-\ln 3 +\sum\limits_{k=2}^n \frac{2}{(2k-3)(2k-1)} \\
&=2-\ln 3 +\sum\limits_{k=2}^n \left(\frac{1}{2k-3}-\frac{1}{2k-1}\right) \\
&=2-\ln 3 +1-\frac{1}{2n-1}<2. \qedhere
\end{align*}
\end{proof}

\begin{proof}[证明二]
下面提供另外一种证明, 注意到不等式(2.7), 我们有
\[ \frac{2}{2k-1}<\frac{1}{2k-2}+\frac{1}{2k}=\frac{1}{2}\left(\frac{1}{k-1}+\frac{1}{k}\right),
\quad (k\geqslant 2). \]
所以
\begin{align*} \sum\limits_{k=1}^n\frac{2}{2k-1}
&<2+\sum\limits_{k=2}^n \frac{1}{2}\left(\frac{1}{k-1}+\frac{1}{k}\right) \\
&=2+\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n-1}\right)+
\frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}\right).
\end{align*}
根据我们先前证明的$\sum\limits_{k=1}^n\frac{1}{k}<1+\ln n$, 于是有
\[ \sum\limits_{k=1}^n\frac{2}{2k-1}<2+\frac{1}{2}[1+\ln (n-1)]+\frac{1}{2}\ln n. \]
于是只需要证明$1+\ln n(n-1)<2\ln (2n+1)$即可, 这容易证明.
\end{proof}

\begin{proof}[证明三]
注意到先前的不等式(\ref{eqgen}), 即
\[ \frac{1}{n+\frac{1}{2}}<\ln \left( 1+\frac{1}{n}\right)=\ln (n+1)- \ln n .\]
所以
\[ \frac{2}{2k-1}=\frac{2}{2(k-1)+1}=\frac{1}{(k-1)+\frac{1}{2}}<\ln k- \ln (k-1),\quad k\geqslant 2. \]
于是有
\[ \sum\limits_{k=1}^n\frac{2}{2k-1}<2+\sum\limits_{k=2}^n[\ln k-\ln (k-1)]=2+\ln n<2+\ln (2n+1).
\qedhere \]
\end{proof}

\begin{proof}[证明四]
不等式等价于
\[ \sum\limits_{k=1}^n\frac{2}{2k-1}<\ln (2n+1). \]
因为我们有不等式$\ln (1+x)<x(x>0)$, 于是待证不等式等价于
\[ \ln \frac{1}{2n+1}<-\sum\limits_{k=2}^n\frac{2}{2k-1}. \]
由于$\ln (1+x)\leqslant x(x>-1)$, 因此有
\[ \ln \frac{2k-1}{2k+1}=\ln \left( 1-\frac{2}{2k+1} \right)<-\frac{2}{2k+1}.\]
所以
\[ \ln \frac{1}{2n+1}
=\sum\limits_{k=1}^n\ln \frac{2k-1}{2k+1}
< -\sum\limits_{k=1}^n \frac{2}{2k+1}
<-\sum\limits_{k=2}^n\frac{2}{2k-1}.\qedhere \]
\end{proof}

\begin{proof}[证明五]
利用不等式$\frac{1}{1+n}<\ln \left( 1+\frac{1}{n}\right)$, 所以有
\[ \frac{2}{2k-1}=\frac{1}{(k-\frac{3}{2})+1}<\ln \left( 1+\frac{1}{k-\frac{3}{2}}\right)=
\ln \frac{2k-1}{2k-3},\quad (k\geqslant 2). \]
于是
\[ \sum\limits_{k=1}^n\frac{2}{2k\!-\!1}\!=\!2\!+\!\sum\limits_{k=2}^n
\frac{2}{2k\!-\!1}\!<\!2\!+\!\sum\limits_{k=2}^n \ln \frac{2k\!-\!1}{2k\!-\!3}
\!=\!2\!+\!\ln (2n\!-\!1)\!<\!2\!+\!\ln (2n\!+\!1). \qedhere \]
\end{proof}

证明五的几何解释.
\begin{center}
\begin{tikzpicture}[domain=-2:4,scale=1.2]
\draw[elegant,color=black,domain=0.3:4.8] plot (\x,{1/((\x))})
node[above ] {$\displaystyle \frac{2}{2x\!-\!1}$};

\draw (0,0) node[below left] {$O$};

\shade[ball color=black](2.35,0.2) circle(0.5pt);
\shade[ball color=black](2.5,0.2) circle(0.5pt);
\shade[ball color=black](2.65,0.2) circle(0.5pt);

\draw[->] (-0.4,0) -- (5,0) node[below=2pt] {$x$};
\draw[->] (0,-0.4) -- (0,3.5) node[left] {$y$};
\end{tikzpicture}
\end{center}
我们保留第一个矩形, 后面的矩形放大为曲线所围面积.
\[ \sum\limits_{k=1}^n\frac{2}{2k-1} <2+\int_1^n \frac{2}{2x-1}\rd x =2+\ln (2n-1)<2+\ln (2n+1). \]

\begin{proof}[证明六]
我们下面证明
\[ \frac{2}{3}+\frac{2}{5}+\frac{2}{7}+\cdots +\frac{2}{2n-1}< \ln (2n+1). \]
这很显然, 因为只需要注意到
\[ \frac{2}{3}+\frac{2}{5}+\cdots +\frac{2}{2n-1}<
\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots +\frac{1}{2n-2}+\frac{1}{2n-1}. \]
这个时候利用不等式(\ref{eq24}), 于是
\[ \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{2n-1}<\ln (2n-1)<\ln (2n+1).\qedhere \]
\end{proof}

\begin{exam}[2013年深圳一模]
已知函数$f(x)=\frac{(x+a)\ln x}{x+1}$, 曲线$y=f(x)$在$(1,f(1))$处的切线与
直线$2x+y+1=0$垂直. \\
\yi 求$a$的值. \\
\er 若任意$x\in [1,+\infty )$, 有$f(x)\leqslant m(x-1)$恒成立, 求$m$的取值范围. \\
\san 求证\sld~ $\ln \sqrt[4]{2n+1}<\sum\limits_{k=1}^n\frac{k}{4k^2-1}$.
\end{exam}

\begin{proof}[证明一]
由(2)可知, 对任意$x\in (1,+\infty )$, 有$ \ln x<\frac{1}{2}(x-\frac{1}{x})$. 令$x=\frac{2k+1}{2k-1}$得
\begin{equation}\label{eq264}
\ln \frac{2k+1}{2k-1}<\frac{1}{2}\left( \frac{2k+1}{2k-1}-\frac{2k-1}{2k+1}\right)
=\frac{4k}{4k^2-1}.\end{equation}
于是
\[ \sum\limits_{k=1}^n\frac{4k}{4k^2-1} >\sum\limits_{k=1}^n \ln \frac{2k+1}{2k-1}
=\ln (2n+1). \qedhere \]
\end{proof}

注意到不等式(\ref{eq264})即为
\[ \frac{1}{2k-1}+\frac{1}{2k+1}>\ln (2k+1)-\ln (2k-1). \]
证明一的几何解释为
\begin{center}
\begin{tikzpicture}[domain=-2:4,scale=2]
\fill[yellow!60!white] (0.5,0) -- (1,0) -- (1,1) -- (0.5,2) -- cycle;
\fill[yellow!60!white] (1,0) -- (1.5,0) -- (1.5,0.6666) -- (1,1) -- cycle;
\fill[yellow!60!white] (1.5,0) -- (2,0) -- (2,0.5) -- (1.5,0.66666) -- cycle;
\fill[yellow!60!white] (2,0) -- (2.5,0) -- (2.5,0.4) -- (2,0.5) -- cycle;

\fill[yellow!60!white] (3.5,0) -- (4,0) -- (4,0.25) -- (3.5,0.30) -- cycle;
%\fill[yellow!60!white] (4,0) -- (4.5,0) -- (4.5,0.2222) -- (4,0.25) -- cycle;

\draw[dashed] (1,0) -- (1,1) -- (0.5,2) --(0.5,0) ;
\draw[dashed] (1.5,0) -- (1.5,0.6666) -- (1,1);
\draw[dashed] (2,0) -- (2,0.5) -- (1.5,0.66666) ;
\draw[dashed] (2.5,0) -- (2.5,0.4) -- (2,0.5) ;

\draw[dashed] (4,0) -- (4,0.25) -- (3.5,0.30) -- (3.5,0);
%\draw[dashed] (4.5,0) -- (4.5,0.2222) -- (4,0.25);

\draw (0.25,0) node[below left] {$O$};

\draw (0.5,0) node[below] {$1$};
\draw (1,0) node[below] {$3$};
\draw (1.5,0) node[below] {$5$};
\draw (2,0) node[below] {$7$};
\draw (3.5,0) node[below] {$2n\!\!-\!\!1$};
\draw (4,0) node[below] {$2n\!\!+\!\!1$};

\shade[ball color=black](2.85,0.2) circle(0.5pt);
\shade[ball color=black](3,0.2) circle(0.5pt);
\shade[ball color=black](3.15,0.2) circle(0.5pt);

\draw[->] (-0.2,0) -- (5,0) node[below=2pt] {$x$};
\draw[->] (0.25,-0.4) -- (0.25,3) node[left] {$y$};
\draw[elegant,color=black,domain=0.4:4.8] plot (\x,{1/((\x))})
node[above ] {$\displaystyle \frac{1}{x}$};
\end{tikzpicture}
\end{center}

\begin{proof}[证明二]
(3)因为
\[ \sum\limits_{k=1}^n \frac{4k}{4k^2-1}=\sum\limits_{k=1}^n
\left(\frac{1}{2k-1} +\frac{1}{2k+1}\right). \]
所以我们只需要证明
\begin{equation}
1+\frac{2}{3}+\frac{2}{5}+\cdots +\frac{2}{2n-1}+\frac{1}{2n+1}>\ln (2n+1).
\end{equation}
注意到不等式
\[ \frac{2}{5}+\frac{2}{7}+\cdots +\frac{2}{2n-1} >
\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\cdots +\frac{1}{2n-1}+\frac{1}{2n}.\]
于是
\[ 1+\frac{2}{3}+\frac{2}{5}+\cdots +\frac{2}{2n-1}+\frac{1}{2n+1}
>1+\frac{2}{3}+\sum\limits_{k=5}^{2n+1}\frac{1}{k} >\frac{5}{3}+\ln (2n+1)-\ln 5, \]
其中最后一个不等式是利用了(\ref{eq2.5}), 接下来很显然有$5/3>\ln 5$.
\end{proof}

证明二的的几何解释为
\begin{center}
\begin{tikzpicture}[domain=-2:4,scale=2]
\fill[yellow!60!white] (0.5,0) -- (0.75,0) -- (0.75,2) -- (0.5,2) -- cycle;
\fill[yellow!60!white] (1,0) -- (1.5,0) -- (1.5,1) -- (1,1) -- cycle;
\fill[yellow!60!white] (1.5,0) -- (2,0) -- (2,0.66666) -- (1.5,0.66666) -- cycle;
\fill[yellow!60!white] (2,0) -- (2.5,0) -- (2.5,0.5) -- (2,0.5) -- cycle;
\fill[yellow!60!white] (3.5,0) -- (4,0) -- (4,0.32) -- (3.5,0.32) -- cycle;
\fill[yellow!60!white] (4,0) -- (4.25,0) -- (4.25,0.27) -- (4,0.27) -- cycle;

\draw[dashed] (0.5,0) -- (0.5,2) --(0.75,2)-- (0.75,0) ;
\draw[dashed] (1,0) -- (1,1) -- (1.5,1) -- (1.5,0.66666);
\draw[dashed] (1.5,0) -- (1.5,0.66666) -- (2,0.66666)--(2,0.5) ;
\draw[dashed] (2,0) -- (2,0.5) -- (2.5,0.5) --(2.5, 0) ;

\draw[dashed] (3.5,0) -- (3.5,0.32) -- (4,0.32)--(4,0.27) ;
\draw[dashed] (4,0) -- (4,0.27) -- (4.25,0.27) -- (4.25,0);

\draw (0.25,0) node[below left] {$O$};

\shade[ball color=black](2.85,0.2) circle(0.5pt);
\shade[ball color=black](3,0.2) circle(0.5pt);
\shade[ball color=black](3.15,0.2) circle(0.5pt);

\begin{exam}[福建某年高考试题]
已知$f(x)=a\ln (x+1)+\frac{1}{x+1}+3x-1$.\\
\yi 若$x\geqslant 0$时, 恒有$f(x)\geqslant 0$成立, 求实数$a$的取值范围. \\
\er 证明\sld~ $\frac{2}{4\times 1^2-1}+\frac{3}{4\times 2^2-1}+\frac{4}{4\times 3^2-1}+\cdots +
\frac{n+1}{4\times n^2-1}>\frac{1}{4}\ln (2n+1)$.
\end{exam}

对于第(2)问的结论, 它比上一个例题要弱, 这里在额外补充两种方法.
\begin{proof}[证明一]
(2) 在(1)中, 我们取$a=-2$, 于是当$x>0$时, 有
\begin{equation} \frac{1}{x+1}+3x-1>2\ln (x+1). \end{equation}
在上式中, 我们令$x=\frac{2}{2k-1}$, 整理即得
\[ \frac{k+1}{4k^2-1}>\frac{1}{4}\ln \frac{2k+1}{2k-1},\quad k=1,2,\ldots .\qedhere \]
\end{proof}

\begin{proof}[证明二]
在不等式$ \ln (1+x)<\frac{x}{\sqrt{1+x}},(x>0)$中, 我们令$x=\frac{2}{2k-1}$, 整理即得
\[ \ln \left(1+\frac{2}{2k-1}\right)<\frac{2}{\sqrt{4k^2-1}}. \]
所以有
\[ \frac{1}{4}\ln \frac{2k+1}{2k-1}<\frac{\frac{1}{2}}{\sqrt{4k^2-1}}
=\frac{\sqrt{k^2-\frac{1}{4}}}{4k^2-1}<\frac{k+1}{4k^2-1}.\qedhere \]
\end{proof}

\begin{exam}
已知函数$f(x)=ax+\frac{b}{x}+2-2a(a>0)$的图像在点$(1,f(1))$处的切线与直线$y=2x+1$
平行. \\
\yi 求$a,b$满足的关系式. \\
\er 若$f(x)\geqslant 2\ln x$在$[1,+\infty )$上恒成立, 求$a$的取值范围. \\
\san 证明\sld~ $1+\frac{1}{3}+\frac{1}{5}+\cdots +\frac{1}{2n-1}>\frac{1}{2}\ln (2n+1)+\frac{n}{2n+1},
(n\in \mathbb{N}^*)$.
\end{exam}

\begin{proof}[证明]
由(2)可知, 对任意$x\geqslant 1$都有
\begin{equation} x-\frac{1}{x}\geqslant 2\ln x. \end{equation}
\end{proof}

\begin{exam}[2011年浙江卷, 理科第22题] \label{2011zhejiangli}
\quad \\
已知函数$f(x)=2a\ln (1+x)-x$, 其中$a>0$. \\
\yi 求$f(x)$的单调区间和极值. \\
\er 求证\sld 对任意$n\in \mathbb{N}^*$有
\[ 4\lg e+\frac{\lg e}{2}+\frac{\lg e}{3}+\cdots +\frac{\lg e}{n} >
\lg \left[ e^{\frac{(1+n)^n}{n^n}} \cdot (n+1)\right].\]
\end{exam}

\begin{proof}[证明]
(2) 不等式等价于
\[ 1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} +3 >\ln (n+1) +
\left( 1+\frac{1}{n}\right)^n. \]
\end{proof}

\section{2007年湖北卷理科}

\begin{exam}[2007年湖北卷, 理科第21题] \label{2007hubeili}
已知$m,n$均为正整数.\\
\yi 用数学归纳法证明\sld 当$x>-1$时, $(1+x)^n>1+nx$.\\
\er 对于$n\geqslant 6$, 已知$\left( 1-\frac{1}{n+3}\right)^n<\frac{1}{2}$, 求证\sld
\[ \displaystyle \Bigl( 1-\frac{m}{n+3}\Bigr)^n<\Bigl( \frac{1}{2}\Bigr)^m,\quad m=1,2,3,\ldots \]
\san 求出满足等式$3^n+4^n+\cdots +(n+2)^n=(n+3)^n$的所有正整数$n$.
\end{exam}

\begin{proof}[证明]
注意到
\[ \Bigl( 1-\frac{m}{n+3}\Bigr)^n\leqslant \Bigl( 1-\frac{1}{n+3}\Bigr)^{mn}<\Bigl( \frac{1}{2}\Bigr)^m.\]
当$n\geqslant 6$时, 很显然有
\begin{gather*}\Bigl( 1-\frac{1}{n+3}\Bigr)^n\!+\Bigl( 1-\frac{2}{n+3}\Bigr)^n\!+\cdots +\Bigl( 1-\frac{n}{n+3}\Bigr)^n\!<\frac{1}{2}+\Bigl( \frac{1}{2}\Bigr)^2\!+\cdots +\Bigl( \frac{1}{2}\Bigr)^n,\\
\Rightarrow \Bigl( \frac{n+2}{n+3}\Bigr)^n+\Bigl( \frac{n+1}{n+3}\Bigr)^n+\cdots +\Bigl( \frac{3}{n+3}\Bigr)^n<1. \end{gather*}
于是只需要验证$n=1,2,3,4,5$, 易知只有$n=2,3$成立.
\end{proof}

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\end{document}