If $2\int_{0}^{1} xf(x)dx\geq\int_{0}^{1}(f(x))^{2}dx$, prove than $\int_{0}^{1}(f(x))^2dx\geq\frac{4}{3}$

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?

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 $$\int_{0}^{1}(f(x))^{2}dx\leq \frac{4}{3}.$$
You need to write that the condition gives $$\int\limits_0^1xf(x)dx\geq0$$.
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: $$\int\limits_0^1f(x)^2dx\leq\frac{4}{3}.$$