How to show the inequality is strict?

This is an exercise from Rudin’s Principles of Mathematical Analysis, Chapter $6$.

Suppose $f$ is a real, continuously differentiable function on $[a, b]$, $f(a) = f(b) = 0$, and
$$\int_a^b f^2(x)\, dx = 1.$$
Prove that
$$\int_a^b xf(x)f'(x)\,dx = -\frac{1}{2}$$
and that
$$\int_a^b [f'(x)]^2\,dx \cdot \int_a^b x^2f^2(x)\,dx {\color{red}>} \frac{1}{4}.$$

The first equality can be easily proved by integrating by parts. Also, I can show that
$$\int_a^b [f'(x)]^2\,dx \cdot \int_a^b x^2f^2(x)\,dx {\color{red}\geq} \frac{1}{4}$$
by Cauchy-Schwarz inequality. However, why the inequality must be always strict? I tried discussing $f'(x) = \lambda xf(x)$ for some constant $\lambda$ when the equality holds and deriving contradiction. But it seems helpless. Thanks in advance for any suggestions!

Solutions Collecting From Web of "How to show the inequality is strict?"

Assume $f'(x)=\lambda xf(x)$. Then $f$ is identically zero.
Indeed assume $f(x_0)\ne 0$ for some $x_0\in [a,b]$. Then $x_0>a$ and we can let $a’$ be the infimum of all numbers such that $f$ has no zero in $(a’,x_0)$.
Then $f(a’)=0$ (this is where we use $f(a)=0$).
In the interval $(a’,x_0)$ we have
$$\frac{\mathrm d}{\mathrm dx}\ln |f(x)|=\frac{f'(x)}{f(x)} =\lambda x$$
and hence $f(x)=c\cdot e^{\frac12\lambda x^2}$. By continuity $0=f(a’)=c\cdot e^{\frac12\lambda a’^2}$ and hence $c=0$ and ultimately $f(x_0)=0$.

The differential equalities gives $f(x)=e^{\lambda \frac{x^2}{2}+c}$ which is never equal to 0 or f constant which can’t satisfy your requests.