How prove this $\int_{a}^{b}[f''(x)]^2dx\ge\dfrac{4}{b-a}$

For the function $$f(x) = \frac{(x-a)(b-x)^2}{(b-a)^2}$$ that integral is exactly $$\frac{4}{b-a}.$$ If $g$ is twice continuously differentiable and $g(a)=g(b)=g'(a)=g'(b)=0$ then

$$\begin{eqnarray} \int_a^b(f''+g'')^2 &\geq& \frac{4}{b-a} +2\int_a^b f'' g''\\ &=& \frac{4}{b-a} - 2\int_a^b f^{(3)}g' \\ &=& \frac{4}{b-a} + 2\int_a^b f^{(4)}g \\ &=& \frac{4}{b-a} \end{eqnarray} $$

This proves the general case since any such function $h$ can be written as $h = f + (h - f)$.

I found $$g(x)=6x-2a-4b$$ because $$\int_{a}^{b}[f''(x)]^2dx\int_{a}^{b}(6x-2a-4b)^2dx\geqslant\left(\int_{a}^{b}(6x-2a-4b)f''(x)dx\right)^2$$

and $$\int_{a}^{b}(6x-2a-4b)^2dx=4(b-a)^3$$ $$\left(\int_{a}^{b}(6x-2a-4b)f''(x)dx\right)^2=[4(b-a)]^2$$

This is the special case of Schoenberg theorem on cubic splines (piecewise polynomials), saying that the cubic spline minimizes the curvature, see these notes, Theorem on p.5.

Since there are no conditions for your function inside the interval, the extremal spline transforms into cubic polynomial. So what you really need is to find cubic polynomial that satisfies boundary conditions.

The proof of Schoenberg theorem is pretty much the same as you outlined in your post.