Integral Inequality $\int\limits_0^1f^2(x)dx\geq12\left( \int\limits_0^1xf(x)dx\right)^2.$ Let $f:[0,1]\to\mathbb{R}$ be a continuous function such that $\int\limits_0^1f(x)dx=0$. Prove that $$\int\limits_0^1f^2(x)dx\geq12\left( \int\limits_0^1xf(x)dx\right)^2.$$
My approach as follow
Let $F(x)=\int\limits_0^xf(t)dt$. 
Integrate by part we have
$$\int\limits_0^1F(x)dx=-\int\limits_0^1xf(x)dx.$$
By Cauchy-Schwarz inequality 
$$\begin{aligned}
\left( \int\limits_0^{1/2}xf(x)dx\right)^2&\leq\left( \int\limits_0^{1/2}x^2dx\right)\left( \int\limits_0^{1/2}f^2(x)dx\right)=\frac{1}{24}\int\limits_0^{1/2}f^2(x)dx\\
\left( \int\limits_{1/2}^1(x-\frac{1}{2})f(x)dx\right)^2&\leq\left( \int\limits_{1/2}^1(x-\frac{1}{2})^2dx\right)\left( \int\limits_{1/2}^1f^2(x)dx\right)=\frac{1}{24}\int\limits_{1/2}^1f^2(x)dx
\end{aligned}$$
Hence $$\left( \int\limits_0^{1/2}xf(x)dx\right)^2+\left( \int\limits_{1/2}^1(x-\frac{1}{2})f(x)dx\right)^2\leq \frac{1}{24}\int\limits_{0}^1f^2(x)dx.$$
But I can't deduce the result. 
Please help me. Thank in advanced.
 A: Since $f$ has mean zero,
$$ \int_{0}^{1} x\,f(x)\,dx = \int_{0}^{1}\left(x-\frac{1}{2}\right)\,f(x)\,dx,\tag{1}$$
and by the Cauchy-Schwarz/Buniakowski inequality we have:
$$ \left(\int_{0}^{1}\left(x-\frac{1}{2}\right)\,f(x)\,dx\right)^2 \leq \int_{0}^{1}f(x)^2\,dx \int_{0}^{1}\left(x-\frac{1}{2}\right)^2\,dx \tag{2}$$
so we just have to check that:
$$ \int_{0}^{1}\left(x-\frac{1}{2}\right)^2\,dx = \frac{1}{12},\tag{3}$$
that is straightforward.
A: A "mechanical" way to deal with this question is to follow the proof 
of Cauchy Inequality and write
\begin{aligned}
0&\leq \int_0^1 (x+\lambda f+\mu)^2\\
 &=E(f^2)\lambda^2+2\lambda E(xf) +\frac{1}{3}+\mu+\mu^2
\end{aligned}
Here $E(g)=\int_0^1 g,\;E^2(g)=(E(g))^2$. Thus 
$$ \Delta_1=4 E^2(xf)-4 E^2(f)\left(\frac{1}{3}+\mu+\mu^2\right)\leq 0$$
Now treat $\Delta_1\leq 0$ as an inequality of $\mu$,
$$\Delta_2=E(f^2)\left (-\frac{1}{3}E(f^2)+4 E^2(xf)\right)\leq 0$$
If $E(f^2)=0$ then $f=0$ almost everywhere, so done. Otherwise, we have
$$ E^2(xf)\leq \frac{1}{12} E(f^2)$$
