设$f(x)$是正的递减的函数，证明

\[ \int_{0}^{1}{xf(x)^2dx} \int_{0}^{1}{f(x)dx}\le \int_{0}^{1}{f(x)^2dx }\int_{0}^{1}{xf(x)dx} \]

证明 考虑

\[ G(x,y)= \frac{1}{2}(x-y)(f(x)-f(y))f(x)f(y)\]

显然有

\[ G(x,y)\leq 0 \]

\[ \iint_{[0,1]^{2}}{G(x,y)dxdy}=\int_{0}^{1}{xf^{2}(x)dx}\cdot \int_{0}^{1}{f(x)dx}-\int_{0}^{1}{xf(x)dx}\cdot\int_{0}^{1}{f^{2}(x)dx}\]

打开即得证