A lower bound in variational inequality Let $f\in C^2([a.b])$, assume that $$f(a)=0=f(b),~f'(a) = 1,f'(b) =0.$$ Prove that $$\int_{a}^{b}\Big|f''(x)\Big|^2\mathrm{d}x\geq \frac{4}{b-a}.$$
Is there a way to convert it into an eigenvalue problem and then one may look for the first eigenvalue as the lower bound?

Change the variables: $x=(b-a)t+a$ and $g(t)=f((b-a)t+a)$
$$\int_a^b|f''(x)|^2dx = \int_0^1 |f''((b-a)t+a)|^2 (b-a) dt =\frac{1}{(b-a)^3} \int_0^1 |g''(t)|^2 dt$$
Therefore the problem is equivalent to proving
$$\int_0^1|g''(t)|^2dt \geq4 (b-a)^2$$
for $g\in C^2([0,1])$ with $g(0)=g(1)=0$, $g'(0)=(b-a)=:c$ and $g'(1)=0$.
Let $\phi$ be smooth then using boundary data and integration by parts and Holder we get $$\left| -c\phi(0)+\int_0^1g\phi'' dx \right|= \left| \int_0^1g''\phi dx \right| \leq \left(\int_0^1|g''|^2\right)^\frac{1}{2} \cdot \left( \int_0^1 \phi^2 \right)^\frac{1}{2}$$ 
Next the problem is to choose the correct $\phi$. Firstly we want $\phi''=0$ so that we get rid of the "bad" term on LHS. So $\phi$ is linear. Also note that the above inequality is invariant under $\phi \mapsto K \cdot \phi$, where $K$ is a constant. So we can look for $\phi(x)=Ax+1$ and find $A$. Now the inequality reads 
$$\int_0^1 |g''|^2 \geq c^2 \frac{|\phi(0)|^2}{\int_0^1 \phi^2}$$
Remains to choose $\phi$ so that the ratio in front of $c^2$ is $4$. So one can find that $\phi(x)=-\frac{3}{2}x+1$ works. Note that $\phi(0)=1$ and
$$\int_0^1 \phi^2 dx = \frac{2}{3} \frac{(\frac{3}{2}x-1)^3}{3} |_0^1 = \frac{1}{4}$$.
Moreover it is the best constant in this family since $\int_0^1 (Ax+1)^2 dx = \frac{1}{3} (A^2+3A+3)$ and we want to minimize this quantity (since it is in the denominator). Minimum achieved at the vertex $A=-\frac{3}{2}$. Also in general the constant $4$ is optimal since if you take $f$ to be $3$rd order polynomial satisfying all the given BC then you obtain equality.
