设 $f(x)$ 在 $[a,b]$ 上连续可微 ($0<a<b$), $f(a)=f(b)=0$, $\dps{\int_a^b f^2(x)\rd x=1}$. 试证: $$\bex \int_a^b x^2f'^2(x)\rd x>\frac{1}{4}.     \eex$$

证明: $$\beex \bea 1&=\sez{\int_a^b f^2(x)\rd x}^2\\ &=\sez{x\cdot f^2(x)|_{x=a}^{x=b} -\int_a^b 2f(x)f'(x)\cdot x\rd x}^2\\ &=\sez{-2\int_a^b f(x)\cdot xf'(x)\rd x}^2\\ &< 4\int_a^b f^2(x)\rd x\cdot \int_a^b x^2f'^2(x)\rd x\\ &=4\int_a^b x^2f'^2(x)\rd x. \eea \eeex$$