How prove this integral inequality $\int_{0}^{1}f^2(x)dx\ge 24\left(\int_{0}^{1}f(x)dx\right)^2$? let $f:[0,1]\longrightarrow R $ be a continuous function, if
$$\int_{0}^{1}x^2f(x)dx=-2\int_{\frac{1}{2}}^{1}F(x)dx$$
where $F(x)=\displaystyle\int_{0}^{x}f(t)dt,x\in [0,1]$,then prove that
$$\int_{0}^{1}f^2(x)dx\ge 24\left(\int_{0}^{1}f(x)dx\right)^2$$
I think this inequality is very interesting, becasue not long ago,I have solve this a little same problem:Funny integral inequality
I believe this two problem have same methods.Thank you everyone
 A: We start with, $\displaystyle \int_{\frac{1}{2}}^1 F(x)\,dx = \int_{\frac12}^1\int_0^x f(t)\,dt \,dx =  \frac{1}{2}\int_0^{\frac12} f(x)\,dx + \int_{\frac12}^1 (1-x)f(x)\,dx$
Thus we have, $$\displaystyle \int_0^1 x^2f(x)\,dx = -\int_0^{\frac12}f(x)\,dx - 2\int_{\frac12}^1 (1-x)f(x)\,dx$$
and, $$\displaystyle \int_0^1 f(x)\,dx = -\int_{\frac12}^1 (x-1)^2f(x)\,dx - \int_0^{\frac12} x^2f(x)\,dx$$
Hence, $\displaystyle \int_{\frac12}^1 \{1+(x-1)^2\}f(x)\,dx + \int_0^{\frac12} \{1+x^2\}f(x)\,dx = 0$
Say, $\phi(x) = 1+x^2 \textrm{ for } x \in [0,\frac12) \textrm{ and } 1+(x-1)^2 \textrm{ for } x \in [\frac12,1]$
Then, $\displaystyle \int_0^1 \phi(x)f(x)\,dx = 0$
Now by Cauchy-Schwarz Inequality :
$\displaystyle \left(\int_0^1 f(x)\,dx \right)^2 = \left(\int_0^1 f(x) + m\phi(x)f(x)\,dx \right)^2 \le \left(\int_0^1 f^2(x)\,dx\right)\left(\int_0^1 (1+m\phi(x))^2\,dx\right)$
Since, $\displaystyle \int_0^1 (1+m\phi(x))^2\,dx = 1 + \frac{13}{6}m +\frac{283}{240}m^2 \ge \frac{4}{849}$
We have, $\displaystyle \left(\int_0^1 f(x)\,dx \right)^2 \le \frac{4}{849}\int_0^1 f^2(x)\,dx$
The constant $\frac{4}{849}$ is the best possible since equality is attained for $f(x) = 1- \frac{260}{283}\phi(x)$.
A: We have
$$
\begin{align}
\int_{1/2}^1F(x)\,\mathrm{d}x
&=\int_{1/2}^1\int_0^xf(t)\,\mathrm{d}t\,\mathrm{d}x\\
&=\color{#00A000}{\int_0^{1/2}\int_{1/2}^1f(t)\,\mathrm{d}x\,\mathrm{d}t}+\color{#C00000}{\int_{1/2}^1\int_t^1f(t)\,\mathrm{d}x\,\mathrm{d}t}\\
&=\int_0^{1/2}\frac12f(t)\,\mathrm{d}t+\int_{1/2}^1(1-t)f(t)\,\mathrm{d}t\\
&=-\frac12\int_0^1t^2f(t)\,\mathrm{d}t\tag{1}
\end{align}
$$
since we are integrating $f(t)$ over the following region in $\mathbb{R}^2$
$\hspace{5cm}$
Rearranging $(1)$, we have
$$
\int_0^{1/2}\frac{t^2+1}{2}f(t)\,\mathrm{d}t+\int_{1/2}^1\frac{(1-t)^2+1}{2}f(t)\,\mathrm{d}t=0\tag{2}
$$
Aside from a scale factor, we want to minimize
$$
\int_0^1f(t)^2\,\mathrm{d}t\tag{3}
$$
given
$$
\int_0^1f(t)\,\mathrm{d}t=1\tag{4}
$$
That means for every variation $\delta f$ so that $(2)$ and $(4)$ hold, that is,
$$
\int_0^{1/2}\frac{t^2+1}{2}\delta f(t)\,\mathrm{d}t+\int_{1/2}^1\frac{(1-t)^2+1}{2}\delta f(t)\,\mathrm{d}t=0\tag{5}
$$
and
$$
\int_0^1\delta f(t)\,\mathrm{d}t=0\tag{6}
$$
we have $(3)$ is stationary, that is,
$$
\int_0^1f(t)\delta f(t)\,\mathrm{d}t=0\tag{7}
$$
Orthogonality says that there must be constants $\mu$ and $\lambda$ so that
$$
f(t)=\mu\cdot1+\lambda\left(\frac{t^2+1}{2}[0\le t\le1/2]+\frac{(1-t)^2+1}{2}[1/2\le t\le1]\right)\tag{8}
$$
Plugging $(8)$ into $(2)$ says
$$
\frac{13}{24}\mu+\frac{283}{960}\lambda=0\tag{9}
$$
Plugging $(8)$ into $(4)$ says
$$
\mu+\frac{13}{24}\lambda=1\tag{10}
$$
Solving $(9)$ and $(10)$ gives
$$
\mu=\frac{849}{4}\qquad\text{and}\qquad\lambda=-390\tag{11}
$$
Plugging $(8)$ and $(11)$ into $(3)$ yields
$$
\int_0^1f(t)^2\,\mathrm{d}t=\frac{849}{4}\tag{12}
$$
Squaring $(4)$, to make things homogeneous, allows us to say
$$
\int_0^1f(t)^2\,\mathrm{d}t\ge\frac{849}{4}\left(\int_0^1f(t)\,\mathrm{d}t\right)^2\tag{13}
$$
