How prove this $\int_{a}^{b}[f''(x)]^2dx\ge\dfrac{4}{b-a}$ let $f$ on $[a,b]$ two  continuously differentiable functions,such 

$$f(a)=f(b)=0, f'(a)=1,f'(b)=0,b>a>0$$
  show that
  $$\int_{a}^{b}[f''(x)]^2dx\ge\dfrac{4}{b-a}$$

My idea: use Cauchy-Schwarz inequality
$$\int_{a}^{b}[f''(x)]^2dx\cdot\int_{a}^{b}g^2(x)dx\ge\left(\int_{a}^{b}f''(x)g(x)dx\right)^2$$
How find the $g(x)$?
then I think 

\begin{align}
\int_{a}^{b}f''(x)g(x)dx &=\int_{a}^{b}g(x)df'(x)=f'(x)g(x)|_{a}^{b}-\int_{a}^{b}f'(x)g(x)dx\\
&=f'(b)g(b)-f'(a)g(a)-\int_{a}^{b}g(x)df(x)\\
&=-g(a)+\int_{a}^{b}g'(x)dF(x)\\
&=-g(a)+F(x)g'(x)|_{a}^{b}-\int_{a}^{b}F(x)g''(x)\\
&=-g(a)+F(b)g'(b)-F(a)g'(a)-\int_{a}^{b}F(x)g''(x)dx
\end{align}
  where $F(x)=\int_{a}^{x}f(t)dt$
Now following I can't find the $g(x)$? can you help me or use other methods solve it? Thank you 

and some hours ago: I ask this integral inequality:How prove this $\int_{a}^{b}f^2(x)dx\le (b-a)^2\int_{a}^{b}[f'(x)]^2dx$
 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 $h = f + (h - f)$.
A: 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
$$\int_{a}^{b}(6x-2a-4b)^2dx=4(b-a)^3$$
$$\left(\int_{a}^{b}(6x-2a-4b)f''(x)dx\right)^2=[4(b-a)]^2$$
A: 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.
A: Let $g(x)$ be a polynomial, by integration by parts,
\begin{align*}
\int_a^bg(x)f''(x)\mathrm{d}x
&=g(x)f'(x)\bigg|_a^b-\int_a^bg'(x)f'(x)\mathrm{d}x\\
&=-g(a)+\underbrace{\int_a^bg''(x)f(x)\mathrm{d}x}_{=0}
\end{align*}
Only let $g''(x)=0\Rightarrow g(x)=c_1x+c_0$. by the Cauchy-Schwarz inequality,
$$\int_a^b\big(f''(x)\big)^2\mathrm{d}x\int_a^bg^2(x)\mathrm{d}x\geqslant\left(\int_a^bg(x)f''(x)\mathrm{d}x\right)^2\longrightarrow\text{Constants}$$
We have,
$$\int_a^b\big(f''(x)\big)^2\mathrm{d}x\geqslant \frac{g^2(a)}{\int_a^bg^2(x)\mathrm{d}x}=\frac{4}{b-a}$$
Substituting,
$$3c_1(b-a)(c_1a+c_0)^2=4\big[(c_1b+c_0)^3-(c_1a+c_0)^3\big]$$
Simplification, we get
$$\big((a+2b)c_1+3c_0\big)^2=0\Rightarrow (a+2b)c_1=-3c_0$$
So, we get,
$$g(x)=c_1x-\frac{1}{3}(a+2b)c_1$$
Inside,$c_1$ is a constant. $\Box$
