TITLE: How prove this integral inequality $\int_{0}^{1}f^2(x)dx\ge 24\left(\int_{0}^{1}f(x)dx\right)^2$?
QUESTION [12 upvotes]: 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

REPLY [2 votes]: Thought：by Cauchy-Schwarz inequality,
$$\bigg(\int_0^1g(x)f(x)\,\mathrm{d} x\bigg)^2\leqslant\int_0^1g^2(x)\,\mathrm{d} x\int_0^1f^2(x)\,\mathrm{d} x$$
We just need to prove that there is a constant K.
$$\int_0^1g(x)f(x)\,\mathrm{d} x=K\int_0^1f(x)\,\mathrm{d} x$$
Proof. by integration by parts,
\begin{align*}
-2\int_{\frac{1}{2}}^{1}F(x)\,\mathrm{d} x&=\int_{\frac{1}{2}}^{1}F(x)\,\mathrm{d}(C-2x)
=(C-2x)F(x)\bigg|_{\frac{1}{2}}^1+\int_{\frac{1}{2}}^{1}(2x-C)f(x)\,\mathrm{d} x\\
&=\int_0^1(C-2)f(x)\,\mathrm{d} x+\int_0^{\frac{1}{2}}(1-C)f(x)\,\mathrm{d} x+\int_{\frac{1}{2}}^{1}(2x-C)f(x)\,\mathrm{d} x
\end{align*}
Since $\textstyle\int_0^1x^2f(x)\,\mathrm{d} x=-2\int_{\frac{1}{2}}^{1}F(x)\,\mathrm{d} x$,
$$\int_0^1(C-2)f(x)\,\mathrm{d} x=\int_0^{\frac{1}{2}}(x^2+C-1)f(x)\,\mathrm{d} x+\int_{\frac{1}{2}}^{1}(C-2x+x^2)f(x)\,\mathrm{d} x$$
Consequently, let
$$g(x)=\begin{cases}
x^2+C-1,& x\in[0,\tfrac{1}{2})\\
C-2x+x^2,& x\in[\tfrac{1}{2},1)
\end{cases}$$
by Cauchy-Schwarz inequality,
\begin{align*}
\left(\int_0^1(C-2)f(x)\,\mathrm{d} x\right)^2&=(C-2)^2\left(\int_0^1f(x)\,\mathrm{d} x\right)^2=\left(\int_0^1g(x)f(x)\,\mathrm{d} x\right)\\
&\leqslant\int_0^1g^2(x)\,\mathrm{d} x\int_0^1f^2(x)\,\mathrm{d} x
\end{align*}
After calculation, we get
$$\int_0^1g^2(x)\,\mathrm{d} x=\frac{240c^2-440c+203}{240}$$
Hence, we have
$$\left(\int_0^1f(x)\,\mathrm{d} x\right)^2\leqslant\frac{240c^2-440c+203}{240(C-2)^2}\int_0^1f^2(x)\,\mathrm{d} x$$
Let $h(C)=\frac{240C^2-440C+203}{240(C-2)^2}\Rightarrow \lim_{C\to2}h(C)=+\infty$
$$h'(C)=\frac{237-260C}{120(C-2)^3} \Leftrightarrow h'(C)=0\Rightarrow C=\frac{237}{260}\Rightarrow h(\tfrac{237}{260})=\frac{4}{849}$$
$$\begin{array}{cccccc}
\hline
 & (-\infty,\tfrac{237}{260}) & \tfrac{237}{260} & (\tfrac{237}{260},2) & (2,+\infty) & +\infty\\\hline
h'(C) & <0 & 0 & >0  & <0 & <0 \\
h(C)  & \downarrow & \text{local minimum} & \uparrow & \downarrow & 1\\\hline
\end{array}$$
So, we have
$$\min_{C\neq2}h(C)=h(\tfrac{237}{260})=\frac{4}{849}$$
Therefore,
$$\int_0^1f^2(x)\,\mathrm{d} x\geqslant\frac{849}{4}\bigg(\int_0^1f(x)\,\mathrm{d} x\bigg)^2$$
Obviously, $\tfrac{849}{4}$ is the best constant, the equality holds if and only if $C=\frac{237}{260}$ and $f(x)=g(x)$