Given $\int_{\frac13}^{\frac23}f(x)dx=0$, how to prove $4860(\int_0^1f(x)dx)^2\le 11\int_0^1|f''(x)|^2dx$? 
Suppose $f\in C^2[0,1]$, and
  $\int_{\frac13}^{\frac23}f(x)dx=0$.
  Prove  that 
  $$\left(\int_0^1f(x)dx\right)^2\le \frac{11}{4860}\int_0^1|f''(x)|^2dx.$$

This problem is quite similar to Prove the following integral inequality: $\int_{0}^{1}(f''(x))^2dx\ge 1920\left(\int_{0}^{1}f(x)dx\right)^2$.
I have tried to write
$$\int_0^1f(x)dx=\int_0^{\frac13}f(x)dx+\lambda \int_{\frac13}^{\frac23}f(x)dx+\int_{\frac23}^{1}f(x)dx$$ for any $\lambda \in \mathbb{R}$, and pick a suitable $g$ such that
$$\int_0^1f(x)dx=\int_0^1g(x)f''(x)dx$$
then we can use Cauchy-Schwarz inequality to get what we want. How can I get the function $g$?
 A: This is essentially following the steps in my answer to a quasi-similar question.
I'm not going to explain how I find the function $g(x)$ below.

Let $X = \mathcal{C}^2[0,1]$ and $P,Q,C : X \to \mathbb{R}$ be functionals over $X$ defined by
$$P(f) = \int_0^1 f''(x)^2 dx,\quad Q(f) = \int_0^1 f(x)dx\quad\text{ and }\quad C(f) = \int_{1/3}^{2/3} f(x) dx$$
The question can be rephrased as

Given $f \in X$ with $C(f) = 0$, how to verify $\;P(f) \ge \frac{4860}{11} Q(f)^2$?

Since both the inequality and constraint is homogeneous in scaling of $f$ by a constant. We can restrict our attention to those $f$ which satisfies $C(f) = 0$ and $Q(f) = 1$.
Consider following functions
$$\phi(x) = x^4 - \frac12 x^2 + \frac{29}{6480}
\quad\text{ and }\quad
\psi(x) = \begin{cases}
\left(\frac13-x\right)^4, & x \le \frac13\\
0, & \frac13 \le x \le \frac23\\
\left(x - \frac23\right)^4, & x \ge \frac23
\end{cases}
 $$
Combine them and define another function $g(x)$ by
$$g(x) = -\frac{405}{11}\left[ \phi\left(x-\frac12\right) - \frac32 \psi(x) \right]$$
It is not hard to check 


*

*$g(x) \in \mathcal{C}^3[0,1] \subset X$.

*$C(g) = 0$, $Q(g) = 1$.

*$g''(0) = g'''(0) = g''(1) = g'''(1) = 0$

*$g''''(x) = \frac{4860}{11}$ for $x \in [0,\frac13)\cup (\frac23,1]$

*$g''''(x) = -\frac{9720}{11}$ for $x \in (\frac13,\frac23)$

*$P(g) = \frac{4860}{11}$.


For any $f \in X$ with $C(f) = 0, Q(f) = 1$, let $\eta = f - g$, we have
$$\begin{align}
  & P(f) - P(g) - P(\eta)\\ 
= & 2\int_0^1 g''(x)\eta''(x) dx\\
= & 2\int_0^1 ( g''(x)\eta'(x))' - g'''(x)\eta'(x) dx\\
= & 2\int_0^1 ( g''(x)\eta'(x) - g'''(x)\eta(x))' + g''''(x)\eta(x)dx\\
= &2\left\{\left[ g''(x)\eta'(x) - g'''(x)\eta(x) \right]_0^1
 + \frac{4860}{11}(Q(\eta)-C(\eta)) -\frac{9720}{11}C(\eta)\right\}
\end{align}
$$
What's in the square bracket vanish because of $(3)$. The remain terms vanish
because 


*

*$Q(\eta) = Q(f) - Q(g) = 1 - 1 = 0$. 

*$C(\eta) = C(f) - C(g) = 0 - 0 = 0$.


Together with the fact $P(\eta)$ is non-negative, we obtain:
$$P(f) = P(g) + P(\eta) \ge P(g) = \frac{4860}{11}$$.
