How prove the integral inequality if $f(x)$ is second-order differentiable and $f(0)=f(1)=f′(1)=0$ $f′(1)=1$ Prove $\int^1_0(f′′(x))^2dx)≥4$ 
I have some trial but failed
 A: For symmetry reasons it does not make a difference whether we assume $$f'(0)=0,\quad f'(1)=1\ ,\tag{1}$$ or the other way around. So I shall assume $(1)$.
Le t $f_0$ be the admissible $f$ for which $$J(f):=\int_0^1\bigl(f''(x)\bigr)^2\ dx$$ is minimal, let $u(\cdot)$ be any function fulfilling
$$u(0)=u(1)=u'(0)=u'(1)=0\ ,\tag{2}$$
and consider the function $f_\epsilon:=f_0+\epsilon u$. Then
$$J(f_\epsilon)=\int_0^1\bigl(f_0''^2+2\epsilon f_0'' u''+ u''^2)\ dx\ .$$
Since $f_0$ is the optimal function we should have
$${d\over d\epsilon}J(f_\epsilon)\biggr|_{\epsilon=0}=0\ .$$
This means $$0=\int_0^1 f_0''\>u''\ dx=\ldots=\int_0^1 f_0^{(4)}(x)u(x)\ dx\ ,$$
and this should be true for all $u$ fulfilling $(2)$. Therefore we conjecture that $f_0$ is a polynomial of degree $\leq3$, and the given boundary conditions then enforce
$$f_0(x)=x^3-x^2\ .\tag{3}$$
In order to prove that this is indeed the optimal function we argue as follows: We can write any $f$ fulfilling the given boundary conditions in the form
$$f(x)=f_0(x)+u(x)$$
with an $u$ satisfying $(2)$ and $f_0$ as in $(3)$. We then obtain
$$\eqalign{\int_0^1 f''^2(x)\ dx&=\int_0^1\bigl((6x-2)^2+(6x-2)u''(x)+u''^2(x)\bigr)\ dx\cr &=\mathstrut\ldots\cr &=\int_0^1\bigl((6x-2)^2+u''^2(x)\bigr)\ dx\geq4\ ,\cr}$$
with equality only if $u''(x)\equiv0$, or $u(x)\equiv0$.
A: First of all, please fix the value of f'(1), because it's both 0 and 1 right now.
Taking f'(0)=1, f'(1)=0 I could not prove the inequality, but I think I made some advancement.
Here is a link to a pdf file of my attempt, hopefully it will help!
https://www.dropbox.com/s/ddz7idzmx5ldbc2/Integral%20Inequality%20Attempt.pdf
