设 $f$ 在 $[a,b]$ 上连续可微且 $f(a)=0$. 试证: $$\bex \int_a^b |f'(x)|^2\rd x\geq \frac{2}{(b-a)^2}\int_a^b |f(x)|^2\rd x. \eex$$

证明: $$\beex \bea f(x)&=\int_a^x f'(t)\rd t,\\ |f(x)|^2&\leq \int_a^x 1^2\rd t\cdot \int_a^x |f'(t)|^2\rd t \leq (a-x)\int_a^b |f'(t)|^2\rd t,\\ \int_a^b |f(x)|^2\rd x &\leq \int_a^b |f'(t)|^2\rd t\cdot \int_a^b (x-a)\rd x =\frac{(b-a)^2}{2} \int_a^b |f'(t)|^2\rd t. \eea \eeex$$