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If $f:[0,1]\rightarrow\mathbb{R}$ is a continuous function such that $2\int_{0}^{1} xf(x)dx\geq\int_{0}^{1}(f(x))^{2}dx$, prove that $\int_{0}^{1}(f(x))^2dx\geq\frac{4}{3}$.

However, using CBS I proved that $\int_{0}^{1}(f(x))^2dx\int_{0}^{1}x^2dx\geq(\int_{0}^{1}(f(x))xdx)^2\geq\frac{1}{4}(\int_{0}^{1}(f(x))^2dx)^2$ and so I obtain exactly the reverse of what I am asked to:$\frac{4}{3}\geq\int_{0}^{1}(f(x))^2dx$. Is there a mistake in the task or in my proof?

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Yes, the inequality $\int_{0}^{1}(f(x))^2dx\geq\frac{4}{3}$ should be reversed. Note that $f=0$ satisfies $$2\int_{0}^{1} xf(x)dx\geq\int_{0}^{1}(f(x))^{2}dx$$ but $\int_{0}^{1}(f(x))^2dx\geq\frac{4}{3}$ does not hold.

On the other hand, by Cauchy-Schwarz (note that $\int_{0}^{1} 2xf(x)dx\geq 0$) $$\left(\int_{0}^{1}(f(x))^{2}dx\right)^2\leq\left(\int_{0}^{1} 2xf(x)dx\right)^2\leq \int_{0}^{1}4x^2dx\int_{0}^{1}(f(x))^2dx=\frac{4}{3}\int_{0}^{1}(f(x))^2dx$$ which implies $$\int_{0}^{1}(f(x))^{2}dx\leq \frac{4}{3}.$$

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You need to write that the condition gives $\int\limits_0^1xf(x)dx\geq0$.

Otherwise, your solution is not full.

Now, by C-S $$\int\limits_0^1f(x)^2dx\int\limits_0^1x^2dx\geq\left(\int\limits_0^1xf(x)dx\right)^2\geq\frac{1}{4}\left(\int\limits_0^1f(x)^2dx\right)^2$$ and since $\int\limits_0^1x^2dx=\frac{1}{3}$, we obtain: $$\int\limits_0^1f(x)^2dx\leq\frac{4}{3}.$$

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