陈洪葛的博客

先看12的。

设$f:[0,1]\rightarrow R$是连续可导函数，若$\displaystyle \int_{0}^{\frac{1}{2}}{f(x)dx}=0 $,求证:
\[ \int_{0}^{1}{f'(x)^{2}dx}\geq 12\left(\int_{0}^{1}{f(x)dx} \right)^{2} \]
(tian 275461)
证明:
\[ \int_{0}^{\frac{1}{2}}{f(x)dx}=0\Rightarrow  \int_{0}^{\frac{1}{2}}{xf'(x)dx}=\frac{1}{2}f\left(\frac{1}{2}\right)\]
\begin{align*}
  \left(\int_{0}^{1}{f(x)dx} \right)^{2}&=\left[\int_{\frac{1}{2}}^{1}{(f(x)-f(\frac{1}{2}))dx}+\frac{1}{2}f(\frac{1}{2}) \right]^{2}\\
&=\left[\int_{\frac{1}{2}}^{1}{\int_{\frac{1}{2}}^{x}{f'(t)  dt}dx}+\int_{0}^{\frac{1}{2}}{xf'(x)dx}  \right]^{2}\\
&=\left[\int_{\frac{1}{2}}^{1}{(1-t)f'(t)dt}+\int_{0}^{\frac{1}{2}}{xf'(x)dx}  \right]^{2}\\   
&\leq 2\left[\int_{\frac{1}{2}}^{1}{(1-t)f'(t)dt}\right]^{2}+2\left[\int_{0}^{\frac{1}{2}}{xf'(x)dx} \right]^{2}\\ 
&\leq 2\left[\int_{\frac{1}{2}}^{1}{(1-t)^{2}dt}\int_{\frac{1}{2}}^{1}{f'(t)^{2}dt}+\int_{0}^{\frac{1}{2}}{x^2dx}\int_{0}^{\frac{1}{2}}{f'(t)^{2}dt} \right]\\
&=\frac{1}{12}\int_{0}^{1}{f'(x)^{2}dx}
  \end{align*}

$\square$

再看27的。

设$f(x)$在$[0,1]$连续可导且可积，若$\displaystyle \int_{\frac{1}{3}}^{\frac{2}{3}}{f(x)dx}=0 $
求证：
\[ \int_{0}^{1}{(f'(x))^{2}dx}\geq 27 \left(\int_{0}^{1}{f(x)dx}\right)^{2} \]

(tian275461)

证明
考虑
\[ G(x)= \left\{
     \begin{array}{ll}
 x, & \hbox{$x\in\left[0,\frac{1}{3}\right)$} \\
 1-2x, & \hbox{$x\in\left[\frac{1}{3},\frac{2}{3}\right]$} \\
 x-1, & \hbox{$x\in\left[\frac{2}{3},1\right)$}
     \end{array}
   \right. \]
由Cauchy-Schwarz容易证明.以下略

3个推广

1.若$f(x):[0,1]\rightarrow R$ 是连续可导函数，且$\displaystyle \int_{\frac{1}{2n}}^{\frac{1}{n}}{f(x)dx}=0 $,则有:
\[ \int_{0}^{1}{(f'(x))^{2}dx}\geq \frac{12n^{2}}{4n^2-10n+7}\left(\int_{0}^{1}{f(x)dx}  \right)^{2} \]
提示：设
\[ g(x)=\left\{
\begin{array}{ll}
 x, & \hbox{$x\in\left[0,\frac{1}{2n} \right]$} \\
 1-(2n-1)x, & \hbox{$x\in\left[\frac{1}{2n},\frac{1}{n} \right]$} \\
 x-1, & \hbox{$x\in\left[\frac{1}{n},1  \right]$.}
\end{array} \right. \]
  然后仿照上面一样用Cauchy-Schwarz
       
2. 若$f(x):[a,b]\rightarrow R$ 是连续可导函数，且$\displaystyle \int_{a}^{b}{f(x)dx}=0 $,则有:
\[ \int_{a}^{2b-a}{(f'(x))^{2}dx}\geq \frac{3}{2(b-a)^{3}}\left(\int_{a}^{2b-a}{f(x)dx}  \right)^{2} \]
提示：设
\[ g(x)=\left\{
\begin{array}{ll}
x-a, & \hbox{$x\in [a,b]$} \\
2b-a-x, & \hbox{$x\in [b,2b-a]$.}
\end{array} \right.\]
        然后仿照上面一样用Cauchy-Schwarz
  3.  若$f(x):[0,1]\rightarrow R$ 是连续可导函数，且$\displaystyle \int_{\frac{1}{2n+1}}^{\frac{2}{2n+1}}{f(x)dx}=0 $,则有:
\[ \int_{0}^{1}{(f'(x))^{2}dx}\geq \frac{3(2n+1)^{2}}{4n^2-6n+3}\left(\int_{0}^{1}{f(x)dx} \right)^{2} \]
提示：设
\[ g(x)=\left\{
\begin{array}{ll}
x, & \hbox{$x\in\left[0,\frac{1}{2n+1} \right]$} \\
1-2nx, & \hbox{$x\in\left[\frac{1}{2n+1},\frac{2}{2n+1} \right]$} \\
x-1, & \hbox{$x\in\left[\frac{2}{2n+1},1  \right]$.}\end{array}\right. \]
  然后仿照上面一样用Cauchy-Schwarz   

设$f(x)$是在$[0,1]$非负的连续的凹函数，且$f(0)=1$, 求证:
\[ 2\int_{0}^{1}{x^2f(x)dx}+\frac{1}{12}\leq \left(\int_{0}^{1}{f(x)dx} \right)^{2} \]
证明
设 \[ F(x)=\int_{0}^{x}{f(t)dt}\]
由于$f(x)$是凹函数，所以有
\[ \frac{f(t)-f(0)}{t-0}\geq \frac{f(x)-f(0)}{x-0}, \qquad  t\in(0,x) \]
\[ \Rightarrow f(t)\geq \frac{t}{x}(f(x)-1)+1 \]
故
\[\begin{align*}
  I=\int_{0}^{1}{x^2f(x)dx}&=\int_{0}^{1}{x^2dF(x)}\\
  &=F(1)-2\int_{0}^{1}{x\int_{0}^{x}{f(t)dt}dx}\\
  &\leq F(1)-I-\frac{1}{3}
\end{align*} \]
所以
\[ 2I\leq F(1)-\frac{1}{3} \]
只要证明
\[ F(1)-\frac{1}{3}+\frac{1}{12}\leq F^{2}(1) \]
\[ \Leftrightarrow \left(F(1)-\frac{1}{4}\right)^{2}\geq 0 \]
显然成立。
$\square$
下面的题和这个有着公共的内核。手段一样，所以在这里不证明了。
若$f:[0,1]\rightarrow \mathbf{R}$是连续凹函数，且满足$f(0)=1,$证明:
\[ \int_{0}^{1}{xf(x)dx}\leq \dfrac{2}{3}\left(\int_{0}^{1}{f(x)dx}\right)^{2} \]

此题由tian275461提供。题目非常困难，仅供观赏(水平一般的同学请勿模仿 )。

求积分
\[ I=\int_{0}^{1}{\frac{\ln{x}}{x^2-x-1}dx} \]
解（tian275461）

由方程$x^2-x-1=0$ 的两个根，为了简单起见，我们记$ r_{1}=\varphi=\dfrac{1+\sqrt{5}}{2},r_{2}=\dfrac{1-\sqrt{5}}{2}=1-\varphi$,
且$r_{1}-r_{2}=\sqrt{5},\varphi^{2}=\varphi+1 , \dfrac{\varphi-1}{\varphi}=\dfrac{1}{\varphi^{2}} $
则有

\[ \begin{align*}
I&=\int_{0}^{1}{\frac{\ln{x}}{x^2-x-1}dx}\\
&=\frac{1}{r_{1}-r_{2}}\int_{0}^{1}{\ln{x}\left(\frac{1}{x-\varphi}-\frac{1}{x-(1-\varphi)} \right)dx}\\
&=\dfrac{1}{\sqrt{5}}\int_{0}^{1}{\ln{x}\left(\frac{1}{x-\varphi}-\frac{\varphi}{\varphi x+1}\right)dx}\\
&=\dfrac{1}{\sqrt{5}}\int_{0}^{1}{\frac{\ln{x}}{x-\varphi}dx}-\frac{\varphi}{\sqrt{5}}\int_{0}^{1}{\frac{\ln{y}}{\varphi y+1}dy}\\
&=\frac{1}{\sqrt{5}}\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{\varphi u}}{u-1}du}-\frac{1}{\sqrt{5}}\int_{0}^{\varphi}{\frac{\ln{\frac{u}{\varphi}}}{u+1}du}\\
&=\frac{\ln{\varphi}}{\sqrt{5}}\left(\int_{0}^{\frac{1}{\varphi}}{\frac{1}{u-1}du}+\int_{0}^{\varphi}{\frac{1}{u+1}du} \right)-\frac{1}{\sqrt{5}}\left(\int_{0}^{\varphi}{\frac{\ln{u}}{1+u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du} \right)\\
&=\frac{\ln{\varphi}}{\sqrt{5}}\ln{\frac{\varphi^{2}-1}{\varphi}}-\frac{1}{\sqrt{5}}\left(\int_{0}^{\varphi}{\frac{\ln{u}}{1+u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du} \right)\\
&=-\frac{1}{\sqrt{5}}\left(\int_{0}^{\varphi}{\frac{\ln{u}}{1+u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du} \right)\\
&=-\frac{1}{\sqrt{5}}\left(\int_{1}^{1+\varphi}{\frac{\ln{(u-1)}}{u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}  \right)\\
&=-\frac{1}{\sqrt{5}}\left(\int_{1}^{\varphi^{2}}{\frac{\ln{(u-1)}}{u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}  \right)\\
&=-\frac{1}{\sqrt{5}}\left(\int_{\frac{1}{\varphi^{2}}}^{1}{\frac{\ln{(1-u)}-\ln{u}}{u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}  \right)\\
&=-\frac{1}{\sqrt{5}}\left(\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}-\ln{(1-u)}}{u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}  \right)\\
&=-\frac{2}{\sqrt{5}}\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}+\frac{1}{\sqrt{5}}\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{(1-u)}}{1-u}du}\\
&=-\frac{2}{\sqrt{5}}\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}-\frac{2}{\sqrt{5}}\ln^{2}{\varphi}\\
&=-\frac{2}{\sqrt{5}}\int_{\frac{1}{\varphi^{2}}}^{1}{\frac{\ln{(1-u)}}{u}du}-\frac{2}{\sqrt{5}}\ln^{2}{\varphi}\\
&=-\frac{2}{\sqrt{5}}\left(\int_{0}^{1}{\frac{\ln{(1-u)}}{u}du}-\int_{0}^{\frac{1}{\varphi^{2}}}{\frac{\ln{(1-u)}}{u}du} \right)-\frac{2}{\sqrt{5}}\ln^{2}{\varphi}\\
&=\frac{\pi^{2}}{3\sqrt{5}}-\frac{2}{\sqrt{5}}\Li_{2}{\left(\frac{1}{\varphi^{2}}\right)}-\frac{2}{\sqrt{5}}\ln^{2}{\varphi}\\
&=\frac{\pi^{2}}{3\sqrt{5}}-\frac{2}{\sqrt{5}}\left(\frac{\pi^{2}}{15}-\ln^{2}{\varphi}
\right)-\frac{2}{\sqrt{5}}\ln^{2}{\varphi}\\
&=\frac{\pi^{2}}{5\sqrt{5}}
\end{align*} \]

$\square$

最后一步用了dilogarithm函数的性质，详细可以参见http://mathworld.wolfram.com/Dilogarithm.html

______________________________________________________________________________

简单一点的做法

我们有
\[ \Li_{2}(x)=-\int_{0}^{x}\frac{\ln(1-t)}{t}dt \]
则
\begin{align*}
\int_{0}^{1}\frac{\ln{x}}{x-a}&=-\frac{1}{a}\cdot\int_{0}^{1}\frac{\ln{x}}{1-\frac{x}{a}}dx\\
&= -\frac{1}{a}\cdot\left(\int_{0}^{1}\frac{\ln\left(\frac{x}{a}\right)}{1-\frac{x}{a}}dx+\ln{a}\int_{0}^{1}\frac{1}{1-\frac{x}{a}}dx\right)\\
&=-\int_{0}^{\frac{1}{a}}\frac{\ln{t}}{1-t}dt-\frac{\ln{a}}{a}\int_{0}^{1}\frac{1}{1-\frac{x}{a}}dx\\
&=-\int_{0}^{\frac{1}{a}}\frac{\ln(1-t)}{t}dt\\
&=\Li_{2}\left(\frac{1}{a}\right)
\end{align*}
注意到
\[ x^2-x-1=\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right) \]
\[ I=\int_{0}^{1}\frac{\ln{x}}{x^2-x-1}dx=\frac{1}{\sqrt{5}}\left(\int_{0}^{1}\frac{\ln{x}}{x-\frac{\sqrt{5}+1}{2}}dx-\int_{0}^{1}\frac{\ln{x}}{x-\frac{1-\sqrt{5}}{2}}dx \right)=\frac{1}{\sqrt{5}}\left(\Li_{2}\left(\frac{\sqrt{5}-1}{2}\right)-\Li_{2}\left(-\frac{\sqrt{5}+1}{2}\right)\right)\]
记$\phi=\frac{\sqrt{5}+1}{2}$,则
\[ I=\frac{1}{\sqrt{5}}\left(\Li_{2}\left(\frac{1}{\phi}\right)-\Li_{2}(-\phi)\right) \]
我们还知道公式
\[ \Li_{2}\left(\frac{1}{\phi}\right)=\frac{1}{10}\pi^2-\ln^{2}\phi \]
\[ \Li_{2}(-\phi)=-\frac{1}{10}\pi^2-\ln^2\phi \]
则
\[ I=\frac{1}{5\sqrt{5}}\pi^2 \]

2011年的圣诞节，越南Le Hai贴出了下面不等式并祝福所有Mathlinker圣诞快乐！

Let $a,\;b,\;c$ be reals such that $a+b+c=0$ and $a^2+b^2+c^2=3.$ Prove that \[{a^5}b + {b^5}c + {c^5}a \le  - 3.\]
 

事实上，用Maple因式分解不难得到它等价于

\[ (a^3+6a^2b+3ab^2-b^3)^2\geq0 \]

只要做点齐次化工作就好。

但过了很久都没看见更理想的证明，直到前不久，tian275461网友给了一个基于三角函数的证明，Can神表示他有Cauchy-Schwarz proof，但他就是不肯说。

Proof (tian 275461)

we use $c=-a-b$put in $a^2+b^2+c^2=3$then  $a^2+ab+b^2=\dfrac{3}{2}$
let $a=x-y,b=x+y,\Longrightarrow c=-2x  $,and $3x^2+y^2=\dfrac{3}{2}$
so we can let
$x=\dfrac{\sqrt{2}}{2}\cos{t},y=\dfrac{\sqrt{6}}{2}\sin{t},t\in(0,2\pi) $
\[\begin{align*}
a^5+b^5c+c^5a&=(x-y)^5(x+y)-2x(x+y)^5-32x^5(x-y)\\
&=\left(\dfrac{\sqrt{2}}{2}\cos{t}-\dfrac{\sqrt{6}}{2}\sin{t}\right)^5\left(\dfrac{\sqrt{2}}{2}\cos{t}+\dfrac{\sqrt{6}}{2}\sin{t}\right)\\
&-2\dfrac{\sqrt{2}}{2}\cos{t}\left(\dfrac{\sqrt{2}}{2}\cos{t}+\dfrac{\sqrt{6}}{2}\sin{t}\right)^5\\
&-32\left(\dfrac{\sqrt{2}}{2}cost\right)^5\left(\dfrac{\sqrt{2}}{2}\cos{t}-\dfrac{\sqrt{6}}{2}\sin{t}\right)\\
&=-\dfrac{3}{8}\left[10-2\sin{\left(6t-\dfrac{\pi}{6}\right)}\right]\le -3
\end{align*} \]

同时，下面推广可以同样证明：

1 if $a+b+c=0,a^2+b^2+c^2=3$,then

\[-\dfrac{3\sqrt{26}}{4}\le a^4b+b^4c+c^4a\le\dfrac{3\sqrt{26}}{4}\]

2 if $a+b+c=0,a^2+b^2+c^2=3$,then
\[-\dfrac{3\sqrt{2}}{2}\le a^2b+b^2c+c^2a\le\dfrac{3\sqrt{2}}{2}\]

3: if $a+b+c=0,a^2+b^2+c^2=3$,then
\[-\dfrac{9\sqrt{38}}{8}\le a^6b+b^6c+c^6a\le\dfrac{9\sqrt{38}}{8}\]

希望以后能看见Can神的CS proof.

。。。。。

2013年3月1日，Vo Quoc Ba Can 终于发了究极的Cauchy-Schwarz proof
Proof. Since $a+b+c=0$ and $a^2+b^2+c^2=3,$ it is easy to obtain the below results:

\[\left\{
  \begin{array}{ll}
    ab+bc+ca=-\dfrac{3}{2}, &  \\
   a^3b+b^3c+c^3a=-(ab+bc+ca)^2=-\dfrac{9}{4}, &\\
   ab^2+bc^2+ca^2+3abc=-(a^2b+b^2c+c^2a), &  \\
   a^3b^3+b^3c^3+c^3a^3=(ab+bc+ca)^3+3a^2b^2c^2=-\dfrac{27}{8}+3a^2b^2c^2, &  \\
   \displaystyle\sum (4ab+2c^2+6bc+3)^2=54, &
  \end{array}
\right.\]

With these results, we have
\begin{align*} {a^5}b + {b^5}c + {c^5}a& = \sum {{a^5}b}\\
 &= \sum {{a^3}b(3 - {b^2} - {c^2})} \\
 &= 3\sum {{a^3}b} - \sum {{a^3}{b^3}} - abc\sum {a{b^2}} \\
  & = - \frac{{27}}{4} + \frac{{27}}{8} - 3{a^2}{b^2}{c^2} - abc\sum {a{b^2}}\\
  &= - \frac{{27}}{8} + abc\sum {{a^2}b} .
\end{align*}
 Therefore, it suffices to prove that
 \begin{equation}
  abc(a^2b+b^2c+c^2a) \le \frac{3}{8}.
 \end{equation}
On the other hand, using the Cauchy-Schwarz inequality, we have
 \[{\left[ {\sum {a(4ab + 2{c^2} + 6bc + 3)} } \right]^2} \le \left( {\sum {{a^2}} } \right)\left[ {\sum {{{(4ab + 2{c^2} + 6bc + 3)}^2}} } \right] = 162.\]
  From this, it follows that
  \[-9\sqrt{2} \le \sum a(4ab+2c^2+6bc+3) \le 9\sqrt{2},\]
   or
   \[-\frac{3}{\sqrt{2}} \le a^2b+b^2c+c^2a+3abc \le \frac{3}{\sqrt{2}}.\]
    The last inequality yields:
    \begin{equation}
      (a^2b+b^2c+c^2a+3abc)^2 \le \frac{9}{2}
    \end{equation}
 Using (2) and the AM-GM inequality, we have
\begin{align*}
 abc(a^2b+b^2c+c^2a) &=\frac{1}{3}\cdot 3abc\cdot (a^2b+b^2c+c^2a)\\ & \le \frac{1}{3} \left(\frac{3abc+a^2b+b^2c+c^2a}{2}\right)^2\le \frac{3}{8},
\end{align*}
      which is (1). So, we are done.

设$x,y,z$为非负数
\[ \sum{\frac{xy}{\sqrt{xy+yz}}}\leq \frac{3\sqrt{3}}{4}\sqrt{\frac{(x+y)(y+z)(x+z)}{x+y+z}} \]
证明：
首先，用$x=bc,y=ac,z=ab$替换，我们可以把不等式写成
\[ \frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\leq \frac{3\sqrt{3}}{4}\sqrt{\frac{(x+y)(y+z)(z+x)}{xy+yz+xz}} \]
由于不等式是cyclic型，故可以设$(y-x)(y-z)\leq 0 $\\
\[ \Leftrightarrow \sum_{cyc}{\frac{x}{\sqrt{(x+y)(x+z)}}\cdot\frac{1}{\sqrt{(x+y)(y+z)}}}\leq \frac{3\sqrt{3}}{4\sqrt{xy+yz+xz}} \]
现在，使用排序不等式
$
LHS\leq \frac{x}{\sqrt{(x+y)(x+z)}}\cdot\frac{1}{\sqrt{(y+z)(z+x)}}+
\frac{y}{\sqrt{(y+z)(y+x)}}\cdot\frac{1}{\sqrt{(y+z)(y+x)}}+\frac{z}{\sqrt{(z+x)(z+y)}}\cdot\frac{1}{\sqrt{(x+y)(x+z)}} $
\[ =\frac{y}{(y+z)(y+x)}+\frac{1}{\sqrt{(y+z)(y+x)}}=\frac{1}{\sqrt{xy+yz+xz}}\cdot\sqrt{\frac{xy+yz+xz}{(y+z)(y+x)}}\cdot\left( 1+\frac{y}{\sqrt{(y+z)(y+x)}} \right ) \]
记$t=\frac{y}{\sqrt{(y+x)(y+z)}}\leq 1 $我们有
\[ \sqrt{\frac{xy+yz+xz}{(y+x)(y+z)}}=\sqrt{1-t^2} \]
使用AM-GM
\[ \frac{1}{\sqrt{xy+yz+xz}}\cdot\sqrt{\frac{xy+yz+xz}{(y+z)(y+x)}}\cdot\left( 1+\frac{y}{\sqrt{(y+z)(y+x)}} \right )=(1+t)\sqrt{1-t^2}\]
\[
=\sqrt{\frac{(1+t)(1+t)(1+t)\cdot 3(1-t)}{3}}\leq \frac{3\sqrt{3}}{4} \]
Done!

$\square$

______________________________________________________________

我想

利用上面那个证明

\[ \sum{\frac{xy}{\sqrt{xy+yz}}}\leq \frac{3\sqrt{3}}{4}\sqrt{\frac{(x+y)(y+z)(x+z)}{x+y+z}}\leq \frac{\sqrt{2}}{2}(x+y+z) \]

应该没有技术上的困难。