积分不等式

设$f''(x)$在$[0,1]$上连续，$f(0)=f(1)=0$，且$\forall x \in (0,1)$，$f(x) \neq 0$，求证：
\[ \int _{0}^{1}\left | \frac{f''(x)}{f(x)} \right |\mathrm{d} x > 4\]

我们去年期末考试题

proof: \\
由于 $ |f(x)|$ 在$ [0,1]$上连续，故必能取到最大值，设$ y_{0}=|f(c)|$
为最大值。于是有：\[ \frac{f(c)-f(0)}{c-0}=\frac{f(c)}{c}=f'(x_{1})
\]
\[ \frac{f(1)-f(c)}{1-c}=\frac{-f(c)}{1-c}=f'(x_{2}) \]
因此，
\[ \int_{0}^{1}{|\frac{f''(x)}{f(x)}| dx}>\int_{x_{1}}^{x_{2}}{|\frac{f''(x)}{f(x)}|
dx}\geq \frac{1}{|f(c)|}|f'(x_{2})-f'(x_{1})|= \frac{1}{c(1-c)}
\]
\\
By AM-GM,we have $ c(1-c) \leq \frac{1}{4} $
\[\int_{0}^{1}{|\frac{f''(x)}{f(x)}| dx}> \frac{1}{4} \]
Done!