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
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 $f(x)=g(x)$