Integral and Inequality

Suppose $f(x)$ is a continuously differentiable function on $[a,b]$ satisfying: $$f(a) = f(b) = 0$$ $$\int_a^b (f(x))^2 dx = 1$$ Then show that: $$\int_a^b x^2 (f'(x))^2 dx \geq \frac{1}{4}$$

Solution

By Cauchy: $$\int_a^b x^2(f'(x))^2 dx = \int_a^b (f(x))^2 dx . \int_a^b x^2(f'(x))^2 dx \geq (\int_a^b xf'(x)f(x)dx)^2$$ The last integral can be evaluated using integration by parts, using $u=x$ and $dv = f'(x)f(x)dx$ which yields $v = (f(x))^2/2$, so that: $$\int_a^b xf'(x)f(x)dx = \frac{bf^2(b) - af^2(a)}{2} - \frac{1}{2} \int_a^b (f(x))^2 dx = -\frac{1}{2}$$ so: $$\int_a^b x^2 (f'(x))^2 dx \geq \frac{1}{4}$$