Let $f$ be a continuous, real-valued function on $[0,1]$.
Define $F$ with $F(0)=f(0)$ and $$F(x)= \frac{1}{x}\int_0^x f(y) dy$$ for $x \in (0,1].$
Show that
$$\int\limits_0^1 F^2(x) dx \leq 4\int\limits_0^1 f^2(x) dx.$$
I think you have to solve the problem with partial integration and the Cauchy-Schwarz inequality.
$$ \int\limits_0^1 \left(\frac{1}{x}\int_0^x f(y) dy\right)^2 dx =\int\limits_0^1 \frac{1}{x^2}\left(\int_0^x f(y) dy\right)^2 dx$$ Partial integration with $ u(x)=x^{-2}, u'(x)=-x^{-1},v(x)=(\int_0^x f(y) dy)^2, v'(x)=2\int_0^x f(y) dy\cdot f(x)\\$ $$ \begin{align} &=-\frac{(\int_0^x f(y) dy)^2}x\Bigg|_0^1-\int_0^1\left(-\frac{2}{x}\int_0^x f(y) dy\cdot f(x)\right)dx\\ &=-F^2(1)+2\int_0^1\frac{\int_0^x f(y) dy}{x}\cdot f(x)dx\\ &\le -F^2(1)+2\,\left(\int_0^1\frac{(\int_0^x f(y) dy)^2}{x^2}dx\int_0^1f^2(x)dx\right)^{1/2}\\ &= -F^2(1)+2\,\left(\int_0^1F^2(x)dx\int_0^1f^2(x)dx\right)^{1/2}\\ \end{align} $$
But I didn't know how to proceed from here.