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