数学人眼中的湖北

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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}

egin{document}

egin{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.

egin{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$b_n=nleft(1+frac{1}{n} ight)^na_n(nin mathbb{N})$,
其中$e$为自然对数的底数.
egin{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
$frac{b_1b_2cdots b_n}{a_1a_2cdots a_n}$的公式, 并给出证明.

item[(3)] 令$c_n=(a_1a_2cdots a_n)^{frac{1}{n}}$, 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
egin{equation} label{e137} sumlimits_{k=1}^nsqrt[k]{a_1a_2cdots a_k}
leqslant esumlimits_{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.

egin{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
egin{align*}sumlimits_{k=1}^nsqrt[k]{a_1a_2cdots a_k}
&=sumlimits_{k=1}^nfrac{sqrt[k]{(a_1b_1)(a_2b_2)cdots (a_kb_k)}}{k+1}
leqslant sumlimits_{k=1}^nfrac{1}{k(k+1)}left( sumlimits_{i=1}^na_ib_i ight) \
&= sumlimits_{i=1}^n a_ib_isumlimits_{j=i}^nfrac{1}{j(j+1)}
= sumlimits_{i=1}^n a_ib_i left( frac{1}{i}-frac{1}{n+1} ight).end{align*}
注意到
[ b_i left( frac{1}{i}-frac{1}{n+1} ight)<frac{b_i}{i}=left( 1+frac{1}{i} ight)^i< e.]
因此, 不等式( ef{e137})成立.

我们接下来说明$e$是最佳的, 令$a_k=1/k$, 于是
[ limlimits_{n o infty}{sumlimits_{k=1}^nfrac{1}{k}}iggm/ {sumlimits_{k=1}^n
frac{1}{sqrt[n]{n!}}}=limlimits_{n o infty}frac{sqrt[n]{n!}}{n}=e.]
所以, $e$是使得不等式( ef{e137})成立的最小常数.
end{proof}

egin{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_2cdots a_k}=frac{1}{sqrt[k]{k!}}cdot sqrt[k]{(1cdot a_1)(2cdot a_2)cdots
(kcdot 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
[ sumlimits_{k=1}^nsqrt[k]{a_1a_2cdots a_k}<ecdot sumlimits_{k=1}^n
frac{a_1+2a_2+cdots +ka_k}{k(k+1)}=ecdot sumlimits_{k=1}^n left(
sumlimits_{j=k}^nfrac{k}{j(j+1)} ight)a_k<esumlimits_{k=1}^na_k.qedhere ]
end{proof}

egin{proof}[证明三]
根据Hardy-Landau不等式(不懂就百度吧), 我们有
[ sumlimits_{k=1}^nleft( frac{a_1^{1/p}+a_2^{1/p}+cdots +a_k^{1/p}}{k} ight)^p
leqslant left( frac{p}{p-1} ight)^psumlimits_{k=1}^na_k.]
我们令$p o +infty$得
[limlimits_{p o +infty}left( frac{a_1^{1/p}+a_2^{1/p}+cdots +a_k^{1/p}}{k} ight)^p
=sqrt[k]{a_1a_2cdots a_k},]
以及
[ limlimits_{p o +infty}left( frac{p}{p-1} ight)^p=e. ]
所以, 不等式( ef{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_2cdots a_n}=frac{sqrt[n]{(lambda_1a_1)(lambda_2a_2)
cdots (lambda_na_n)}}{sqrt[n]{lambda_1lambda_2cdots lambda_n}}
leqslant frac{frac{1}{n}sumlimits_{k=1}^nlambda_ka_k}{sqrt[n]{
lambda_1lambda_2cdots lambda_n}}.]
特别地, 取$lambda_k=k$时有
[ sqrt[n]{a_1a_2cdots 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}sumlimits_{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( ef{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.

egin{prop}
设$a_1,a_2,ldots ,a_n$为非负实数, $e$为自然对数的底数, 则
egin{equation}label{e138} sumlimits_{k=1}^nsqrt[k]{a_1a_2cdots a_k}
leqslant esumlimits_{k=1}^na_k - nsqrt[n]{a_1a_2cdots a_n}.end{equation}
end{prop}

egin{proof}[证明]
利用归纳法就等价于证明
[ ea_n+(n-1)sqrt[n-1]{a_1a_2cdots a_{n-1}}geqslant (n+1)sqrt[n]{a_1a_2cdots a_n}.]
由AM-GM不等式可得
[ ea_n+(n-1)sqrt[n-1]{a_1a_2cdots a_{n-1}}geqslant nsqrt[n]{ea_1a_2cdots a_n}.]
再结合不等式$e>left( 1+frac{1}{n} ight)^n$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.

egin{prop}[羊明亮]
设$x_1,x_2,ldots ,x_n$为任意实数, 证明sld
[ sumlimits_{k=1}^nleft( frac{1}{k}sumlimits_{j=1}^kx_j ight)^2leqslant
sumlimits_{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.
[ sumlimits_{k=1}^nleft( frac{1}{k}sumlimits_{j=1}^kx_j ight)^2 leqslant sumlimits_{k=1}^n(k+1)x_k^2 - frac{1}{n}left(sumlimits_{k=1}^nx_k ight)^2.]

egin{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{sumlimits_{j=1}^nsqrt[j]{a_1a_2cdots a_j}}{sumlimits_{j=1}^na_j} ight)^{!!1/n}
+frac{sqrt[n]{a_1a_2cdots a_n}}{sumlimits_{j=1}^nsqrt[j]{a_1a_2cdots a_j}}leqslant
frac{n+1}{n}.]
end{prop}

oindent
{f 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.

egin{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.
[ sumlimits_{k=1}^nfrac{k}{sumlimits_{j=1}^kfrac{1}{a_j}}leqslant Csumlimits_{k=1}^na_k.]
end{prop}

egin{proof}[证明]
根据Cauchy-Schwarz不等式可得
[ left( sumlimits_{j=1}^kfrac{1}{a_j} ight)left( sumlimits_{j=1}^kj^2a_j ight)geqslant
left( sumlimits_{j=1}^kj ight)^2=left[frac{k(k+1)}{2} ight]^2.]
整理便有
[ frac{k}{sumlimits_{j=1}^kfrac{1}{a_j}}leqslant frac{4}{k(k+1)^2}left( sumlimits_{j=1}^kj^2a_j ight).]
所以
egin{align*}
sumlimits_{k=1}^nfrac{k}{sumlimits_{j=1}^kfrac{1}{a_j}}
&leqslant sumlimits_{k=1}^nfrac{4}{k(k+1)^2}left( sumlimits_{j=1}^kj^2a_j ight) \
&=2sumlimits_{j=1}^nj^2a_jsumlimits_{k=j}^nfrac{2}{k(k+1)^2}\
&<2sumlimits_{j=1}^nj^2a_jsumlimits_{k=j}^nleft( frac{1}{k^2}-frac{1}{(k+1)^2} ight) \
&=2sumlimits_{j=1}^nj^2a_jleft[ frac{1}{j^2}-frac{1}{(n+1)^2} ight] \
&<2sumlimits_{j=1}^na_j.end{align*}
因此, 可取$C=2$, 下面说明$2$是最佳系数. 令$a_j=1/j(j=1,2,cdots,n)$,
[ sumlimits_{k=1}^nfrac{k}{sumlimits_{j=1}^kfrac{1}{a_j}}=sumlimits_{k=1}^n
frac{k}{frac{1}{2}k(k+1)}=2sumlimits_{k=1}^nfrac{1}{k+1}leqslant Csumlimits_{k=1}^nfrac{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:
egin{exam}[AMM, 11145] label{exam1321}
设$x_1,x_2,ldots ,x_n$均为正数, 试证明sld
[ sumlimits_{k=1}^nfrac{k}{x_1+x_2+cdots +x_k}
leqslant 2sumlimits_{k=1}^nfrac{1}{x_k}. ]
end{exam}

oindent
{f 注}~
从该不等式可以看出\
(1) 若$x_k>0$, 且$sumlimits_{n=1}^{infty}frac{1}{x_n}$收敛,
则级数
$ sumlimits_{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.
[ sumlimits_{k=1}^nfrac{2k+1}{x_1+x_2+cdots +x_k}
leqslant 4sumlimits_{k=1}^nfrac{1}{x_k}.]

egin{prop}
设$a_kgeqslant 0(k=1,2,ldots ,n)$, 则对每个正整数$m$有
[ sumlimits_{k=1}^nsqrt[k]{a_1a_2cdots a_k}leqslant frac{1}{m}sumlimits_{k=1}^n
a_kleft( frac{k+m}{k} ight)^k.]
end{prop}

section{2014年湖北卷理科}

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

egin{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$eln pi $, $pi $与$3ln pi$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$eln pi >e(2-frac{e}{pi })>2.7 imes (2-0.88)=3.024>3$, 即$pi^e >e^3$. 另一方面,
$3ln pi >3(2-frac{e}{pi })>3 imes (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}

egin{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}

egin{exam}[2011年湖北卷, 理科第21题] label{2011hubeili}quad \
(1) 已知函数$f(x)=ln x-x+1,xin (0,+infty)$, 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$a_1b_1+a_2b_2+cdots+a_nb_nleqslant b_1+b_2+cdots +b_n$, 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}

egin{proof}[证明]
(2) 根据Jensen不等式有
[frac{sumlimits_{k=1}^nb_kcdot ln a_k}{sumlimits_{k=1}^nb_k}leqslant ln left(frac{sumlimits_{k=1}^nb_kcdot a_k}{sumlimits_{k=1}^nb_k} ight)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.

egin{thm} label{thmthm}
对于任意的$xgeqslant -1$, 下述不等式成立,
egin{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,
egin{equation}label{eq2}
frac{1}{1+y}< ln left(1+frac{1}{y} ight)<frac{1}{y} ,
ext{~其中$y>0$或$yleqslant -1$}.end{equation}
end{thm}

egin{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( ef{eq1})的一个直接应用是证明如下常见不等式:
egin{gather} a<frac{a-b}{ln a-ln b}<b,quad 0<a<b.
end{gather}
对于不等式( ef{eq2}), 变形即有,
egin{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
egin{equation} label{eq2.5}
left{ egin{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 &cdotscdots \
ln (n+1) -ln n <&frac{1}{n}<ln n -ln (n-1).
end{aligned}
ight.
end{equation}
将上面所有不等式累加可得
egin{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).
egin{gather}
frac{ln 2}{2}cdot frac{ln 3}{3}cdotfrac{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( ef{eq24})的几何直观如下.
egin{center}
egin{tikzpicture}[domain=-2:4,scale=1.2]
draw[elegant,color=black,domain=0.3:4.8] plot (x,{1/((x))})
node[above ] {$displaystyle frac{1}{x}$};

fill[yellow!60!white] (0,0) -- (0.5,0) -- (0.5,2) -- (0,2) -- cycle;
fill[yellow!60!white] (0.5,0) -- (1,0) -- (1,1) -- (0.5,1) -- cycle;
fill[yellow!60!white] (1,0) -- (1.5,0) -- (1.5,0.66666) -- (1,0.66666) -- cycle;
fill[yellow!60!white] (1.5,0) -- (2,0) -- (2,0.5) -- (1.5,0.5) -- cycle;
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);
draw[dashed] (1.5,0) -- (1.5,0.66666) -- (1,0.66666) ;
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) ;
draw[dashed] (4,0) -- (4,0.23) -- (3.5,0.23) ;

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

draw (0.5,0) node[below] {$1$};
draw (1,0) node[below] {$2$};
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}
egin{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] {$O$};

draw (0.5,0) node[below] {$1$};
draw (1,0) node[below] {$2$};
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](2.85,0.2) 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 ef{thmthm}的加强.
egin{thm}
(1) 当$xin (0,+infty)$时, 有$displaystyle frac{x}{1+frac{1}{2}x}<ln (1+x)<frac{x}{sqrt{1+x}}$.\
(2) 当$xin (-1,0)$时, 有$displaystylefrac{x}{sqrt{1+x}}<ln (1+x)<frac{x}{1+frac{1}{2}x}$.
end{thm}

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

该定理的几何直观如下:\
egin{center}
egin{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
egin{equation}label{eqgen}
frac{1}{y+frac{1}{2}}<ln left( 1+frac{1}{y} ight)=ln (y+1)- ln y<frac{1}{sqrt{y(y+1)}}
<frac{1}{2}left( frac{1}{y}+frac{1}{y+1} ight).
end{equation}
上式的几何解释如下面左图.
egin{center}
egin{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}
egin{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] (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] {$O$};

draw (0.5,0) node[below] {$1$};
draw (1,0) node[below] {$2$};
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](2.85,0.2) 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}

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

egin{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
[ limlimits_{n o infty}
left[ 1!+!frac{1}{2}!+!frac{1}{3}!+!cdots !+!frac{1}{n}!-! ln (n!+!1) ight]
!!=!limlimits_{n o infty }
left[ 1!+!frac{1}{2}!+!frac{1}{3}!+!cdots !+!frac{1}{n}!-! ln n ight]
!!:=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} ight). ]
于是
[ frac{1}{2}left( frac{1}{k} -frac{1}{k+1} ight)
<frac{1}{k}-Bigl[ ln (k+1)-ln kBigr] <
2left( frac{1}{2k}-frac{1}{2k+1} ight). ]
将上式$k=1,2,ldots ,n$求和可得
[ frac{1}{2}left( 1 -frac{1}{n+1} ight)<
y_n < 2 left( frac{1}{2}-frac{1}{3} +frac{1}{4}-frac{1}{5} + cdots +
frac{1}{2n}-frac{1}{2n+1} ight). ]
在上式中令$n o 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{1cdot 2}}+frac{1}{sqrt{2cdot 3}}+frac{1}{sqrt{3cdot 4}}+cdots +
frac{1}{sqrt{ncdot (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
egin{equation}label{eq257}
frac{2}{3}<ln 2<frac{sqrt{2}}{2}. end{equation}

egin{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$nge 1$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.
egin{center}
egin{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] (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] {$O$};

draw (0.5,0) node[below] {$1$};
draw (1,0) node[below] {$2$};
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](2.85,0.2) 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}

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

类似的问题还有许多, 下面简单列举几例.

egin{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$xin [0,+infty )$, 有$f(x)leqslant kx^2$成立, 求实数$k$的最小值. \
san 证明sld~ $displaystyle sumlimits_{k=1}^nfrac{2}{2k-1}-ln (2n+1)<2$.
end{exam}

egin{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 xgeqslant 0. ]
令$g'(x)=0$, 得到$x_1=0,x_2=frac{1-2k}{2k}>-1$. \
(i) 注意到$0leqslant x-ln (x+1)leqslant kx^2$, 于是$kleqslant 0$不符合题意. \
(ii) 当$kgeqslant 1/2$时, $x_2leqslant 0$, 于是在$[0,+infty )$上恒有$g'(x)leqslant 0$. 因此,
对任意的$xin [0,+infty )$时, 有$g(x)leqslant g(0)=0$, 即$f(x)leqslant kx^2$在$[0,+infty )$上恒成立,
故$kgeqslant 1/2$符合题意. \
(iii) 当$0<k<1/2$时, 有$frac{1-2k}{2k}>0$, 对于$xin (0,frac{1-2k}{2k} )$时, $g'(x)geqslant 0$,
$g(x)$在$(0,frac{1-2k}{2k} )$上单调递增, 于是当$xin (0,frac{1-2k}{2k} )$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$ngeqslant 2$时, 注意到
[ ln (2n+1)=sumlimits_{k=1}^nln frac{2k+1}{2k-1}=sumlimits_{k=1}^n ln
left(1+frac{2}{2k-1} ight). ]
于是,
[ sumlimits_{k=1}^nfrac{2}{2k-1}-ln (2n+1)=sumlimits_{k=1}^n
left[ frac{2}{2k-1}-ln left(1+frac{2}{2k-1} ight) ight]. ]
在第(2)小问中, 我们有$x-ln (1+x)leqslant kx^2(kgeqslant frac{1}{2})$, 于是取$k$最小有
egin{equation} x-ln (1+x) leqslant frac{1}{2}x^2. end{equation}
从而, 当$kgeqslant 2$时, 有
[ frac{2}{2k-1}-ln left(1+frac{2}{2k-1} ight)leqslant frac{1}{2}left(frac{2}{2k-1} ight)^2
=frac{2}{(2k-1)^2}<frac{2}{(2k-3)(2k-1)}. ]
因此
egin{align*}
sumlimits_{k=1}^nfrac{2}{2k-1}-ln (2n+1)
&<2-ln 3 +sumlimits_{k=2}^n frac{2}{(2k-3)(2k-1)} \
&=2-ln 3 +sumlimits_{k=2}^n left(frac{1}{2k-3}-frac{1}{2k-1} ight) \
&=2-ln 3 +1-frac{1}{2n-1}<2. qedhere
end{align*}
end{proof}

egin{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} ight),
quad (kgeqslant 2). ]
所以
egin{align*} sumlimits_{k=1}^nfrac{2}{2k-1}
&<2+sumlimits_{k=2}^n frac{1}{2}left(frac{1}{k-1}+frac{1}{k} ight) \
&=2+frac{1}{2}left(1+frac{1}{2}+frac{1}{3}+cdots +frac{1}{n-1} ight)+
frac{1}{2}left(frac{1}{2}+frac{1}{3}+cdots +frac{1}{n} ight).
end{align*}
根据我们先前证明的$sumlimits_{k=1}^nfrac{1}{k}<1+ln n$, 于是有
[ sumlimits_{k=1}^nfrac{2}{2k-1}<2+frac{1}{2}[1+ln (n-1)]+frac{1}{2}ln n. ]
于是只需要证明$1+ln n(n-1)<2ln (2n+1)$即可, 这容易证明.
end{proof}

egin{proof}[证明三]
注意到先前的不等式( ef{eqgen}), 即
[ frac{1}{n+frac{1}{2}}<ln left( 1+frac{1}{n} ight)=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 kgeqslant 2. ]
于是有
[ sumlimits_{k=1}^nfrac{2}{2k-1}<2+sumlimits_{k=2}^n[ln k-ln (k-1)]=2+ln n<2+ln (2n+1).
qedhere ]
end{proof}

egin{proof}[证明四]
不等式等价于
[ sumlimits_{k=1}^nfrac{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}<-sumlimits_{k=2}^nfrac{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} ight)<-frac{2}{2k+1}.]
所以
[ ln frac{1}{2n+1}
=sumlimits_{k=1}^nln frac{2k-1}{2k+1}
< -sumlimits_{k=1}^n frac{2}{2k+1}
<-sumlimits_{k=2}^nfrac{2}{2k-1}.qedhere ]
end{proof}

egin{proof}[证明五]
利用不等式$frac{1}{1+n}<ln left( 1+frac{1}{n} ight)$, 所以有
[ frac{2}{2k-1}=frac{1}{(k-frac{3}{2})+1}<ln left( 1+frac{1}{k-frac{3}{2}} ight)=
ln frac{2k-1}{2k-3},quad (kgeqslant 2). ]
于是
[ sumlimits_{k=1}^nfrac{2}{2k!-!1}!=!2!+!sumlimits_{k=2}^n
frac{2}{2k!-!1}!<!2!+!sumlimits_{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.
egin{center}
egin{tikzpicture}[domain=-2:4,scale=1.2]
draw[elegant,color=black,domain=0.3:4.8] plot (x,{1/((x))})
node[above ] {$displaystyle frac{2}{2x!-!1}$};

fill[yellow!60!white] (0,0) -- (0.5,0) -- (0.5,2) -- (0,2) -- cycle;
fill[yellow!60!white] (0.5,0) -- (1,0) -- (1,1) -- (0.5,1) -- cycle;
fill[yellow!60!white] (1,0) -- (1.5,0) -- (1.5,0.66666) -- (1,0.66666) -- cycle;
fill[yellow!60!white] (1.5,0) -- (2,0) -- (2,0.5) -- (1.5,0.5) -- cycle;
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);
draw[dashed] (1.5,0) -- (1.5,0.66666) -- (1,0.66666) ;
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) ;
draw[dashed] (4,0) -- (4,0.23) -- (3.5,0.23) ;

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

draw (0.5,0) node[below] {$1$};
draw (1,0) node[below] {$2$};
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}
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.
[ sumlimits_{k=1}^nfrac{2}{2k-1} <2+int_1^n frac{2}{2x-1} d x =2+ln (2n-1)<2+ln (2n+1). ]

egin{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( ef{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}

egin{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$xin [1,+infty )$, 有$f(x)leqslant m(x-1)$恒成立, 求$m$的取值范围. \
san 求证sld~ $ln sqrt[4]{2n+1}<sumlimits_{k=1}^nfrac{k}{4k^2-1}$.
end{exam}

egin{proof}[证明一]
由(2)可知, 对任意$xin (1,+infty )$, 有$ ln x<frac{1}{2}(x-frac{1}{x})$. 令$x=frac{2k+1}{2k-1}$得
egin{equation}label{eq264}
ln frac{2k+1}{2k-1}<frac{1}{2}left( frac{2k+1}{2k-1}-frac{2k-1}{2k+1} ight)
=frac{4k}{4k^2-1}.end{equation}
于是
[ sumlimits_{k=1}^nfrac{4k}{4k^2-1} >sumlimits_{k=1}^n ln frac{2k+1}{2k-1}
=ln (2n+1). qedhere ]
end{proof}

注意到不等式( ef{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
egin{center}
egin{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}

egin{proof}[证明二]
(3)因为
[ sumlimits_{k=1}^n frac{4k}{4k^2-1}=sumlimits_{k=1}^n
left(frac{1}{2k-1} +frac{1}{2k+1} ight). ]
所以我们只需要证明
egin{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}+sumlimits_{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( ef{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
egin{center}
egin{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] {$O$};

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}

egin{exam}[福建某年高考试题]
已知$f(x)=aln (x+1)+frac{1}{x+1}+3x-1$.\
yi 若$xgeqslant 0$时, 恒有$f(x)geqslant 0$成立, 求实数$a$的取值范围. \
er 证明sld~ $frac{2}{4 imes 1^2-1}+frac{3}{4 imes 2^2-1}+frac{4}{4 imes 3^2-1}+cdots +
frac{n+1}{4 imes n^2-1}>frac{1}{4}ln (2n+1)$.
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.
egin{proof}[证明一]
(2) 在(1)中, 我们取$a=-2$, 于是当$x>0$时, 有
egin{equation} frac{1}{x+1}+3x-1>2ln (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}

egin{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} ight)<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}

egin{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$f(x)geqslant 2ln x$在$[1,+infty )$上恒成立, 求$a$的取值范围. \
san 证明sld~ $1+frac{1}{3}+frac{1}{5}+cdots +frac{1}{2n-1}>frac{1}{2}ln (2n+1)+frac{n}{2n+1},
(nin mathbb{N}^*)$.
end{exam}

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

egin{exam}[2011年浙江卷, 理科第22题] label{2011zhejiangli}
quad \
已知函数$f(x)=2aln (1+x)-x$, 其中$a>0$. \
yi 求$f(x)$的单调区间和极值. \
er 求证sld 对任意$nin mathbb{N}^*$有
[ 4lg 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) ight].]
end{exam}

egin{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} ight)^n. ]
end{proof}

section{2007年湖北卷理科}

egin{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$ngeqslant 6$, 已知$left( 1-frac{1}{n+3} ight)^n<frac{1}{2}$, 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}

egin{proof}[证明]
注意到
[ Bigl( 1-frac{m}{n+3}Bigr)^nleqslant Bigl( 1-frac{1}{n+3}Bigr)^{mn}<Bigl( frac{1}{2}Bigr)^m.]
当$ngeqslant 6$时, 很显然有
egin{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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L. Maligranda,
Why H"{o}lder's inequality should be called Rogers' inequality,
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ibitem{LS05}
Y.-C. Li and S.-Y. Shaw,
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M. Shao, extit{Proof without words: Bounding the Euler-Mascheroni constant},
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ibitem{HK07}
胡克, 解析不等式的若干问题. 武汉: 武汉大学出版社, 2007.3.

end{thebibliography}
end{document}