Proving $\frac 1 4 < \left( \int_a^b [f']^2 \ dx\right)^{1/2}\left( \int_a^b x^2[f(x)]^2 \ dx\right)^{1/2}$ In chapter 6 of Rudin's Principles of Mathematical Analysis, problem 15, he writes

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
$$ \int_a^b x f(x) f'(x) \ dx = -\frac 1 2 $$
and that
$$\int_a^b[f'(x)]^2 \ dx \cdot \int_a^b x^2f^2(x)\ dx &gt; \frac 1 4. $$

I've shown everything up to demonstrating
$$\int_a^b[f'(x)]^2 \ dx \cdot \int_a^b x^2f^2(x)\ dx \ge \frac 1 4 $$
However, I can't figure out how to eliminate the case where equality holds.  I've seen several other resources say that equality would imply
$$ f'(x) = \lambda xf(x) $$
but I can't figure out why that would be implied.
Resources I've already looked at: How to show the inequality is strict?
Baby Rudin Chapter 6, Problem 15 : Strict inequality
Prob. 15, Chap. 6, in Baby Rudin: If $f$ is a real, continuously differentiable function on $[a, b]$, . . .
Proving a strict inequality (Application of H&#246;lder&#39;s Inequality)
Show that if $f(a)=f(b)=0$ and $\int_a^b [f(x)]^2dx=1$, then $\int_a^b [f&#39;(x)]^2dx\cdot\int_a^b [xf(x)]^2dx\gt \frac14$
I'll rederive Cauchy-Schwarz inequality (specifically the equality condition) for continuous functions on an interval. Suppose $f,g : [a,b] \to \mathbb{R}$ are continuous. Throughout the following, all integrals are over $[a,b]$. Since $f$ and $g$ are continuous on a compact set, they are bounded, and so $\int f^2, \int g^2, \int fg$ all exist. I'll assume that $\int f^2 &gt; 0$. I'll also assume that you can show the following result - if $h$ is continuous on an interval $[a,b]$, then $\int_{[a,b]} h^2 = 0$ if and only if $h$ is identically zero on the interval. [This is why I can assume $\int f^2 &gt; 0$ - otherwise $f$ is the $0$ function and there's nothing to show]
Consider the function $$ J(t) := \int (t f - g)^2 = t^2 \int f^2 - 2 t \int fg + \int g^2.$$
We can derive Cauchy-Schwarz by observing that $J(t) \ge 0$ since it is the integral of a non-negative function, and then minimising the quadratic. Notice that as long as $\int f^2 &gt; 0,$ the quadratic has a unique minima at $t_* = \int fg/ \int f^2$.  So we have that for all $t \in \mathbb{R},$ $$ J(t) \ge J(t_*) = \int g^2 -  \frac{(\int fg)^2}{\int f^2} \ge 0.$$
Now, we want to argue that if $(\int fg)^2 = \int f^2 \int g^2,$ then there must exist a $t \in \mathbb{R}$ such that $tf = g.$ But notice that $(\int fg)^2 = \int f^2 \int g^2$ implies that for $t_* = \int fg/ \int f^2,$  $J(t_*) = 0$. Due to the definition of $J$, this means that $\int (t_*f - g)^2 = 0.$ But $t_* f - g$ is continuous, so this means that $t_* f = g$ everywhere on the interval.
Now you can apply this equality condition to the continuous functions $xf$ and $f'$.
Use Cauchy Schwartz inequality, $p(x)$ and $q(x)$ are continuous in $[a,b]$, then
$$\left(\int_{a}^{b} p(x) q(x) dx \right)^{2} \le \int_{a}^{b} p^2(x) dx \int_{a}^{b} q^2(x) dx,$$
by choosing $p(x)=f'(x), q(x)=x f(x).$ Equality holds when $p(x)=t q(x)$.
Here the equality will hold if $$txf(x)=f'(x) \implies f(x)=C e^{tx^2/2}.$$
Using $f(a)=0=f(b)$, we get $C=0$. so the equality will hold for the trivial  case of $f(x)=0.$