问题：设\(\displaystyle f\left( x \right)\)在\(\displaystyle \left( 0,1 \right)\)上二阶可导，\(\displaystyle f‘‘\left( x \right) >0\)，\(\displaystyle f\left( 0 \right) =0\)，求证：
\[\int_0^1{xf\left( x \right) \text{d}x}\geqslant \frac{2}{3}\int_0^1{f\left( x \right) \text{d}x}
\]
过程如下：由于\(\displaystyle f‘‘\left( x \right) >0\)，所以有
\[\frac{f\left( x \right) +f\left( 0 \right)}{2}\geqslant \frac{\int_0^x{f\left( t \right) \text{d}t}}{x}
\]
即
\[xf\left( x \right) \geqslant 2\int_0^x{f\left( t \right) \text{d}t}
\]
对上式左右两边同时对\(\displaystyle x\)在0到1上积分，得到
\[\begin{align*}
\int_0^1{xf\left( x \right) \text{d}x}&\geqslant 2\int_0^1{\left( \int_0^x{f\left( t \right) \text{d}t} \right) \text{d}x}
\&=2x\int_0^x{f\left( t \right) \text{d}t}\mid_{0}^{1}-2\int_0^1{xf\left( x \right) \text{d}x}
\&=2\int_0^1{f\left( x \right) \text{d}x}-2\int_0^1{xf\left( x \right) \text{d}x}
\end{align*}
\]
移项从而得到
\[\int_0^1{xf\left( x \right) \text{d}x}\geqslant \frac{2}{3}\int_0^1{f\left( x \right) \text{d}x}
\]
【14】利用函数的凹凸性证明一道积分不等式
原文：https://www.cnblogs.com/xuke0721/p/14945868.html
