积分不等式

\begin{liti}
(2022年山大考研)设函数$f$在$[0,1]$上有二阶连续导数,且$f(0)=f(1)=f'(0)=0,f'(1)=1$.求证:

$$
\int_0^1{\left( f''\left( x \right) \right) ^2dx}\geqslant 4,
$$
并指出不等式中等号成立的条件.
\end{liti}
\begin{jieda}
作三次多项式$p(x)=x^3-x^2$,它也满足$p(0)=p(1)=p'(0)=0,p'(1)=1$.此时,
$$
\int_0^1{\left( p''\left( x \right) \right) ^2dx}=\int_0^1{\left( 6x-2 \right) ^2dx}=\left[ \frac{1}{18}\left( 6x-2 \right) ^3 \right] _{0}^{1}=4.
$$

由于
\begin{align*}
\int_0^1{f''\left( x \right) \left( 6x-2 \right) dx} &=\int_0^1{\left( 6x-2 \right) df'\left( x \right)}\\
&=4f'\left( 1 \right) +2f'\left( 0 \right) +6\int_0^1{f'\left( x \right) dx}\\
&=4f'\left( 1 \right) +2f'\left( 0 \right) +6\left[ f\left( 1 \right) -f\left( 0 \right) \right] =4,
\end{align*}
于是
\begin{align*}
\int_0^1{\left( f''\left( x \right) \right) ^2dx}+\int_0^1{\left( p''\left( x \right) \right) ^2dx} &=\int_0^1{\left( f''\left( x \right) -p''\left( x \right) \right) ^2dx}+2\int_0^1{f''\left( x \right) p''\left( x \right) dx}\\
&=\int_0^1{\left( f''\left( x \right) -p''\left( x \right) \right) ^2dx}+2\int_0^1{f''\left( x \right) \left( 6x-2 \right) dx}\\
&=\int_0^1{\left( f''\left( x \right) -p''\left( x \right) \right) ^2dx}+8,
\end{align*}
则
$$
\int_0^1{\left( f''\left( x \right) \right) ^2dx}=\int_0^1{\left( f''\left( x \right) -p''\left( x \right) \right) ^2dx}+4\ge 4,
$$
当且仅当$f''=p''$,即$f(x)=p(x)=x^3-x^2$时取等号成立.
\end{jieda}

 

\begin{liti}
设$f(x)$在区间$[0,1]$上具有连续导数,且$f(0)=0,f(1)=1$.
证明:
$$
\int_0^1{\left| f'\left( x \right) -f\left( x \right) \right|dx}\ge \frac{1}{e}.
$$
\end{liti}
\begin{jieda}
由于
$$
f'\left( x \right) -f\left( x \right) =e^x\left[ e^{-x}f\left( x \right) \right] '
$$
且$e^x\geqslant 1,x\in [0,1]$,故
\begin{align*}
\int_0^1{\left| f'\left( x \right) -f\left( x \right) \right|dx} &=\int_0^1{\left| e^x\left[ e^{-x}f\left( x \right) \right] ' \right|dx}\\
&\geqslant \int_0^1{\left[ e^{-x}f\left( x \right) \right] 'dx}=\left. e^{-x}f\left( x \right) \right|_{0}^{1}=\frac{f\left( 1 \right)}{e}-f\left( 0 \right) =\frac{1}{e}.
\end{align*}
\end{jieda}

\begin{liti}
给定严格递增的无界正实数列$\{a_n\}$,求证:

(1)存在自然数$k_0$,使得对一切正整数$k\geqslant k_0$,均有
$\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots+\frac{a_k}{a_{k+1}}<k-1$;

(2)存在自然数$k_1$,使得对一切正整数$k\geqslant k_1$,均有$\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots+\frac{a_k}{a_{k+1}}<k-2014$.
\end{liti}
\begin{jieda}

\end{jieda}

\begin{liti}
已知数列$\{a_n\}$满足$a_1=2,a_{n+1}=a_n^2-a_n+1$.

(1)求证:当正整数$n\geqslant 2$时有$1-\frac{1}{2^{2^{n-1}}}
<\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}<1-\frac{1}{2^{2^{n}}}$.

(2)对于正整数数列$\{b_n\}$,满足对任意$n$, $\frac{1}{b_1}+\frac{1}{b_2}+\cdots+\frac{1}{b_n}<1$.
证明: $\frac{1}{b_1}+\frac{1}{b_2}+\cdots+\frac{1}{b_n}\leqslant \frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}$.
\end{liti}
\begin{jieda}

\end{jieda}

 

\begin{liti}
无穷正实数数列$\{x_n\}$满足$x_0=1,x_{n+1}\leqslant x_n\ (n=0,1,2,\cdots)$.

(1)求证:对于满足上述要求的任意数列$\{x_n\}$,存在正整数$n$,使得$\frac{x_0^2}{x_1}+\frac{x_1^2} {x_2}++\frac{x_{n-1}^2} {x_n}\geqslant 3.999$;

(2)构造满足要求数列$\{x_n\}$,使得任意正整数$n$, $\frac{x_0^2} {x_1}+\frac{x_1^2} {x_2}++\frac{x_{n-1}^2}{x_n}<4$.
\end{liti}
\begin{jieda}

\end{jieda}

 

\begin{liti}
已知数列$\{a_n\}$满足$a_1=1,a_na_{n+1}=n+1\ (n=1,2,\cdots)$.求证:
$$
\sum_{k=1}^n{\frac{1}{a_k}}\geqslant 2\left( \sqrt{n+1}-1 \right).
$$
\end{liti}
\begin{jieda}

\end{jieda}

 

\begin{liti}(北京市海淀区2021届上学期高三年级期末练习)
设$A$是由$n\times n\ (n\geqslant 2)$个实数组成的$n$行$n$列的数表,满足:每个数的绝对值是$1$,且所有数的和是非负数,则称数表$A$是“$n$阶非负数表”.

(I)判断如下数表$A_1,A_2$是否是“$4$阶非负数表”;

\begin{minipage}{\textwidth}
\begin{minipage}[t]{0.45\textwidth}
\centering
\begin{tabular}{|c|c|c|c|}\hline
1 & 1 &$-1$ &$-1$\\ \hline
1 & 1 &$-1$ &$-1$\\ \hline
1 & $-1$ &1 &$-1$\\ \hline
1 & 1 &$-1$ &$-1$\\ \hline
\end{tabular}
\makeatletter\def\@captype{table}\makeatother\caption{数表$A_1$}
\end{minipage}
\begin{minipage}[t]{0.45\textwidth}
\centering
\begin{tabular}{|c|c|c|c|}\hline
$-1$ & $-1$ &$-1$ &$-1$\\ \hline
1 & 1 &1 &$-1$\\ \hline
1 & $-1$ &1 &$-1$\\ \hline
1 & 1 &$-1$ &$-1$\\ \hline
\end{tabular}
\makeatletter\def\@captype{table}\makeatother\caption{数表$A_2$}
\end{minipage}
\end{minipage}

(II)对于任意“5阶非负数表”$A$,记$R(s)$为$A$的第$s$行各数之和($1\leqslant s\leqslant 5$),证明:存在$\{i,j,k\}\subseteq\{1,2,3,4,5\}$,使得$R(i)+R(j)+R(k)\geqslant 3$;

(III)当$n=2k\ (k\in \mathbb{N}^\ast)$时,证明:对于任意“$n$阶非负数表”$A$,均存在$k$行$k$列,使得这$k$行$k$列交叉处的$k^2$个数之和不小于$k$.
\end{liti}
\begin{jieda}
记$a(i,j)$为数表$A$中第$i$行第$j$列的数, $\sum_{i=1}^{n}\sum_{j=1}^{n}a(i,j)$为数表$A$中所有数的和, $\sum_{i=1}^{k}\sum_{j=1}^{k}a(i,j)$为数列$A$中前$k$行$k$列交叉处各数之和.

(I) $A_1$是“$4$阶非负数表”; $A_2$不是“$4$阶非负数表”.

(II)证明:由题意知$a(i,j)\in\{1,-1\},i=1,2,3,4,5,j=1,
2,3,4,5$,且数表$A$是“$5$阶非负数表”,所以$R(s)\ (s=1,2,3,4,5)$为奇数,且$R(1)+R(2)+R(3)+R(4)+R(5)\geqslant 0$.

不妨设$R(1)\geqslant R(2)\geqslant R(3)\geqslant R(4)\geqslant R(5)$.

\ding{172}当$R(3)\geqslant 0$时,因为$R(3)$为奇数,所以$R(3)\geqslant 1$.
所以$R(1)+R(2)+R(3)\geqslant 3R(3) \geqslant 3$.

\ding{173}当$R(3)<0$时,因为$R(3)$为奇数,所以$R(3)\leqslant -1$.
所以$R(4)+R(5)\leqslant 2R(3)\leqslant -2$.

所以$R(1)+R(2)+R(3)\geqslant -R(4)-R(5)\geqslant 2$.
又因为$R(1),R(2),R(3)$均为奇数,所以$R(1)+R(2)+R(3)\geqslant 3$.

(III)证明: (1)先证明数表$A$中存在$n-1$行$n$列($n=2k$),其所有数的和大于等于$0$.

设$R(i)=\sum_{j=1}^{n}a(i,j)\ (i=1,2,\cdots,n)$,由题意知$\sum_{i=1}^{n}R(i)\geqslant 0$.

不妨设$R(1)\geqslant R(2)\geqslant\cdots\geqslant R(n)$.

由于$n\sum_ {i=1}^{n-1}R(i)-(n-1)\sum_ {i=1}^{n}R(i)=\sum_ {i=1}^{n-1}R(i)-(n-1)\cdot R(n)\geqslant 0$,所以$\sum_ {i=1}^{n-1}R(i)\geqslant \frac{n-1}{n}\sum_{i=1}^{n}R(i)\geqslant 0$.

(2)由(1)及题意不妨设数表$A$前$n-1$行$n$列($n=2k$),其所有数的和大于等于$0$.

下面考虑前$2k-1$行,证明存在$2k-1$行$k$列,其所有数的和大于等于$k$.

设$T(j)=\sum_{i=1}^{2k-1}a(i,j)\ (j=1,2,\cdots,2k)$,则$\sum_{j=1}^{2k}T(j)=\sum_{i=1}^{2k-1}R(i)\geqslant 0$.

不妨设$T(1)\geqslant T(2)\geqslant\cdots\geqslant T(2k)$.

因为$T(j)$为$2k-1$个奇数的和,所以$T(j)$为奇数($j=1,2,\cdots,2k$).

\ding{172}当$T(k)\geqslant 0$时,因为$T(k)$为奇数,所以$T(k)\geqslant 1$,所以$\sum_{j=1}^{k}T(j)\geqslant kT(k)\geqslant k$.

\ding{173}当$T(k)<0$时,因为$T(k)$为奇数,所以$T(k)\leqslant -1$,

所以$\sum_{j=k+1}^{2k}T(j)\leqslant kT(k)\leqslant -k$,所以$\sum_{j=1}^{k}T(j)\geqslant -\sum_{j=k+1}^{2k}T(j)\geqslant k$.

(3)在(2)所设数表$A$下,证明前$2k-1$行前$k$列中存在$k$行$k$列,其所有数的和大于等于$k$.

设$R'(i)=\sum_ {j=1}^{k}a(i,j)\ (i=1,2,\cdots,2k-1)$,则$\sum_ {i=1}^{2k-1}R'(i)=\sum_{j=1}^{k}T(j)\geqslant k$.

不妨设$R'(1)\geqslant R'(2)\geqslant\cdots\geqslant R'(2k-1)$.

\ding{172}当$R'(k)\geqslant 1$时, $\sum_{i=1}^{k}R'(i)\geqslant kR'(k)\geqslant k$;

\ding{173}当$R'(k)\leqslant 0$时, $R'(2k-1)\leqslant R'(2k-2)\leqslant\cdots\leqslant R'(k)\leqslant 0$,所以$\sum_{i=1}^{k}R'(i)\geqslant k-\sum_ {i=k+1}^{2k-1}R'(i)\geqslant k$,所以$\sum_{i=1}^{k}\sum_{j=1}^{k}a(i,j)=\sum_{i=1}^{k}R'(i)\geqslant k$.

综上所述,对于任何“$n$阶非负数表”$A$,均存在$k$行$k$列,使得这$k$行$k$列交叉处的所有数之和不小于$k$.
\end{jieda}

2013-东城区一模