设 $f(x)$ 在 $[a,b]$ 上一阶连续可导, $f(a)=0$. 证明: $$\bex \int_a^b f^2(x)\rd x\leq \cfrac{(b-a)^2}{2}\int_a^b [f'(x)]^2\rd x -\cfrac{1}{2}\int_a^b [f'(x)]^2 (x-a)^2\rd x. \eex$$
证明: $$\beex \bea \int_a^b f^2(x)\rd x &=\int_a^b \sez{\int_a^xf'(t)\rd t}^2\rd x\\ &\leq \int_a^b \sez{ \int_a^x f'^2(t)\rd t \cdot \int_a^x 1^2\rd t }\rd x\\ &=\int_a^b \int_a^x (x-a)f'^2(t)\rd t\rd x\\ &=\int_a^b \int_t^b (x-a)f'^2(t)\rd x\rd t\\ &=\int_a^b f'^2(t)\int_t^b (x-a)\rd x\rd t\\ &=\int_a^b f'^2(t)\cfrac{(b-a)^2-(t-a)^2}{2}\rd t. \eea \eeex$$
