设$f:[0,1]\rightarrow R$是连续函数，且$\displaystyle \int_{0}^{1}{f^{3}(x)dx}=0$,求证:
\[ \int_{0}^{1}{f^{4}(x)dx}\geq \frac{27}{4}\left(\int_{0}^{1}{f(x)dx} \right)^{4} \]

(tian276461)
证明:令
 \[ I_{n}=\int_{0}^{1}{f^{n}(x)dx} \]
由Cauchy-Schwarz得
\[ I_{2}\geq I_{1}^{2} \]
又由Cauchy-Schwarz得
\[ \left(\int_{0}^{1}{(r+f^{2}(x))\cdot f(x)dx} \right)^{2}\leq \int_{0}^{1}{(r+f^{2}(x))^{2}dx}\cdot\int_{0}^{1}{f^{2}(x)dx} \]
展开得
\[ (I_{2}-I^{2}_{1})r^2+2I_{2}^{2}r+I_{2}I_{4}\geq 0 \]
上式恒成立，所以 $ \Delta\leq 0 $
\[ \Rightarrow I_{4}\geq \frac{I^{3}_{2}}{I_{2}-I_{1}^{2}} \]
所以只要证明
\[ \frac{I^{3}_{2}}{I_{2}-I^{2}_{1}}\geq \frac{27}{4}I^{4}_{1} \]
由AM-GM
\[ (I_{2}-I_{1}^{2})I^{4}_{1}=\frac{1}{2}(2I_{2}-2I_{1}^{2})\cdot I^{2}_{1}\cdot I^{2}_{1}\leq \frac{4}{27}I_{2}^{3}\]
故有
\[ \int_{0}^{1}{f^{4}(x)dx}\geq \frac{27}{4}\left(\int_{0}^{1}{f(x)dx} \right)^{4} \]
$\square$