## 数学人眼中的湖北（五）

**数学人眼中的湖北（五）**

(2015年湖北卷, 理科第22题)已知数列$\{a_n\}$的各项均为正数, 且$\displaystyle b_n=n\left(1+\frac{1}{n}\right)^na_n(n\in \mathbb{N})$,其中$e$为自然对数的底数.

(1) 求函数$f(x)=1+x-e^x$的单调区间, 并比较$\displaystyle \left(1+\frac{1}{n}\right)^n$与$e$的大小.

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

(3) 令$\displaystyle c_n=(a_1a_2\cdots a_n)^{\frac{1}{n}}$, 数列$\{a_n\}$和$\{c_n\}$的前$n$项和分别记为
$S_n,T_n$, 试证明: $T_n<eS_n$.

\begin{equation*} \label{e137} \sum\limits_{k=1}^n\sqrt[k]{a_1a_2\cdots a_k}

**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{equation*}\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) \end{equation*}
\begin{equation*}= \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{equation*}

\begin{equation*} b_i \left( \frac{1}{i}-\frac{1}{n+1}\right)<\frac{b_i}{i}=\left( 1+\frac{1}{i}\right)^i< e.\end{equation*}

\begin{equation*} \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.\end{equation*}

**证明二.**

\begin{equation*}\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}.\end{equation*}

\begin{equation*} \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(

**证明三.**

\begin{equation*}\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.\end{equation*}

\begin{equation*}\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},\end{equation*}

\begin{equation*} \lim\limits_{p\to +\infty}\left( \frac{p}{p-1}\right)^p=e. \end{equation*}

**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.

\begin{equation*}\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}}.\end{equation*}

\begin{equation*}\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.\end{equation*}

**命题1.** 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*}

**证明.** 利用归纳法就等价于证明
\begin{equation*} 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}.\end{equation*}

\begin{equation*} ea_n+(n-1)\sqrt[n-1]{a_1a_2\cdots a_{n-1}}\geqslant n\sqrt[n]{ea_1a_2\cdots a_n}.\end{equation*}

**命题2.** (羊明亮)设$x_1,x_2,\ldots ,x_n$为任意实数, 证明:
\begin{equation*}\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{equation*}

**注:** 同样地, 利用归纳法可以证明如下不等式.
\begin{equation*}\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.\end{equation*}

**命题3.** (2005年国家队选拔赛试题)

\begin{equation*}\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{equation*}

**注:** 该不等式是Carleman不等式的加强(为什么).

**命题4.** 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.
\begin{equation*} \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{equation*}

**证明.** 根据Cauchy-Schwarz不等式可得
\begin{equation*}\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.\end{equation*}

\begin{equation*} \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).\end{equation*}

\begin{equation*}\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) \end{equation*}
\begin{equation*}=2\sum\limits_{j=1}^nj^2a_j\sum\limits_{k=j}^n\frac{2}{k(k+1)^2}\end{equation*}
\begin{equation*}<2\sum\limits_{j=1}^nj^2a_j\sum\limits_{k=j}^n\left( \frac{1}{k^2}-\frac{1}{(k+1)^2}\right) \end{equation*}
\begin{equation*}=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{equation*}

\begin{equation*} \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}.\end{equation*}

**例.** (AMM, 11145)

\begin{equation*} \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{equation*}

**注.**

(1) 若$x_k>0$, 且$\displaystyle \sum\limits_{n=1}^{\infty}\frac{1}{x_n}$收敛,

$\displaystyle \sum\limits_{n=1}^{\infty}\frac{n}{x_1+x_2+\cdots +x_n}$

(2)下面这个不等式留给读者.
\begin{equation*} \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}.\end{equation*}

**命题5.**

\begin{equation*} \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{equation*}

posted on 2020-04-09 15:48  Eufisky  阅读(386)  评论(0编辑  收藏  举报