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This is Romanian mathematical competitions 2006 $12^{th}$ grade problem 67:

Prove that for any continuous function $f:[0,1]\to \mathbb{R}$ $$\dfrac{4}{15}\int_{0}^{1}f^2(x)dx\ge\int_{0}^{1}f(x)dx\int_{0}^{1}x^4f(x)dx\tag{4}$$ Also,find the cases of equality.

Now only prove following inequality $$\int_{0}^{1}f^2(x)dx\ge 3\int_{0}^{1}f(x)dx\int_{0}^{1}x^4f(x)d\tag{3}x$$

Because by Cauchy-Schwarz inequality we have $$ \begin{split} \int_{0}^{1}f^2(x)dx\int_{0}^{1}x^8dx &\ge \left(\int_{0}^{1}x^4f(x)dx\right)^2 \quad\text{(1)}\\ \int_{0}^{1}f^2(x)dx\int_{0}^{1}1dx &\ge \left(\int_{0}^{1}f(x)dx\right)^2 \quad\text{(2)} \end{split} $$ $(1)\times (2)$ we have (3) hold,but How to prove (4)?

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Let $$A=\int_{0}^{1}f(x)dx, \quad B=\int_{0}^{1}x^{4}f(x)dx, \quad C=\int_{0}^{1}f(x)^{2}dx.$$

For any $a, b\in \mathbb{R}$, we have $$\int_{0}^{1}(a+bx^{4}-f(x))^{2}dx\geq 0$$ which is equivalent to $$ a^{2}+\frac{2ab}{5}+\frac{1}{9}b^{2}-2aA-2bB+C\geq 0. $$ By differentiating this equation w.r.t $a$ and $b$, as a function of $a$ and $b$ it takes minimum at $$ a=\frac{25A-45B}{16}, \quad b=\frac{-45A+225B}{16}. $$ Put this back into the previous equation and we get $$ C\geq \frac{25}{16}A^{2}-\frac{45}{8}AB+\frac{225}{16}B^{2}=\left(\frac{5}{4}A-\frac{15}{4}B\right)^{2}+\frac{15}{4}AB\geq \frac{15}{4}AB. $$ equality holds for $A=3B$, i.e. $a=(15/8)B, b=(45/8)B, f(x)=a+bx^{4}=c(1+3x^{4})$ for some $c\in \mathbb{R}$.

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