设$f(x)$是$[0,1]\rightarrow R$上的连续函数，且记$\displaystyle F(x)=\int_{0}^{x}{f(t)dt}$,并有
\[ \int_{0}^{1}{x^2f(x)dx}=-2\int_{\frac{1}{2}}^{1}{F(t)dt} \]
求证
\[\int_{0}^{1}{f^2(x)dx}\geq 80\left( \int_{0}^{1}{f(x)dx}\right)^2\]
(tian_275461)

 首先由条件，我们看到
 \[  \int_{0}^{1}{x^2F'(x)dx}=x^2F(x)\bigg|_{0}^{1}-2\int_{0}^{1}{xF(x)dx}=F(1)-2\int_{0}^{1}{xF(x)dx}\]
 \[ \Rightarrow\qquad F(1)=2\int_{0}^{1}{xF(x)dx}-2\int_{\frac{1}{2}}^{1}{F(t)dt}=2\left(\int_{0}^{\frac{1}{2}}{xF(x)dx}+\int_{\frac{1}{2}}^{1}{(x-1)F(x)dx}  \right)\]
 而要证的不等式等价于
 \[ \int_{0}^{1}{(F'(x))^{2}dx}\geq 24(F(1))^2=80\left[2\left(\int_{0}^{\frac{1}{2}}{xF(x)dx}+\int_{\frac{1}{2}}^{1}{(x-1)F(x)dx}  \right)\right]^{2} \]
 注意到由Cauchy-Schwarz不等式，我们有
 \begin{align*}
 \int_{0}^{\frac{1}{2}}{x^4} \int_{0}^{\frac{1}{2}}{(F'(x))^{2}dx}&\geq \left(\int_{0}^{\frac{1}{2}}{x^2F'(x)dx}\right)^{2}\\
 &=\left[\frac{1}{4}F\left(\frac{1}{2}\right)-2\int_{0}^{\frac{1}{2}}{xF(x)dx}\right]^{2}\\
 &=A^2
 \end{align*}
 \begin{align*}
 \int_{\frac{1}{2}}^{1}{(x-1)^4}\int_{\frac{1}{2}}^{1}{(F'(x))^{2}dx}&\geq  \left(\int_{\frac{1}{2}}^{1}{(x-1)^2F'(x)dx}\right)^{2}\\
 &= \left[-\frac{1}{4}F\left(\frac{1}{2}\right)-2\int_{\frac{1}{2}}^{1}{(x-1)F(x)dx}\right]^{2}\\
 &=B^2
 \end{align*}
 而由Cauchy-Schwarz不等式，我们有
 \[A^2+B^2\geq \frac{1}{2} (A+B)^2=\frac{1}{2} F^{2}(1) \]
 而
 \[ \int_{0}^{\frac{1}{2}}{x^4dx}=\int_{\frac{1}{2}}^{1}{(1-x)^4dx}=\frac{1}{160}\]
 故
 \[ \int_{0}^{1}{f^2(x)dx}\geq 80\left( \int_{0}^{1}{f(x)dx}\right)^2\]