TITLE: How prove this $\int_{a}^{b}[f''(x)]^2dx\ge\dfrac{4}{b-a}$
QUESTION [11 upvotes]: 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 $\int_{a}^{b}f^2(x)dx\le (b-a)^2\int_{a}^{b}[f'(x)]^2dx$

REPLY [2 votes]: 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. $\Box$