【14】利用函数的凹凸性证明一道积分不等式

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：

\[\int_0^1{xf\left( x \right) \text{d}x}\geqslant \frac{2}{3}\int_0^1{f\left( x \right) \text{d}x} \]

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过程如下：由于\(\displaystyle f''\left( x \right) >0\)，所以有

\[\frac{f\left( x \right) +f\left( 0 \right)}{2}\geqslant \frac{\int_0^x{f\left( t \right) \text{d}t}}{x} \]

\[xf\left( x \right) \geqslant 2\int_0^x{f\left( t \right) \text{d}t} \]

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

\[\int_0^1{xf\left( x \right) \text{d}x}\geqslant \frac{2}{3}\int_0^1{f\left( x \right) \text{d}x} \]