Let $f:[a,b] \to \mathbb{R}$ be a differentiable function with $f(a)=0$ and so that $f'$ is continuous. Show that $$\int_a^b |{f(x)f'(x)}|\:\mathrm{d}x \leq \frac{b-a}{2} \int_a^b f'(x)^2\: \mathrm{d}x.$$
I received a hint for examining the function $G(x):=\int_a^x |f'(t)|\:\mathrm{d}t$ but haven't really gotten anywhere with it. Any tips would be greatly appreciated
Maybe too long for a comment, did you tried Holder's inequality ? $$\int_a^b|f(x)f'(x)|\mathrm{d}x \le \left( \int_{a}^{b}|f(x)|^2 \mathrm{d}x \right)^{\frac{1}{2}}\left( \int_{a}^{b}|f'(x)|^2 \mathrm{d}x \right)^{\frac{1}{2}}\tag{1}$$ Now write $$f(x)=\int_{a}^{x}f'(t)\mathrm{d}t \tag{1}$$ Now from (2) we may write $$|f(x)| \le \int_a^x|f'(t)|\mathrm{d}t \le (x-a)^{\frac{1}{2}}\left( \int_a^x |f'(t)|^2 \mathrm{d}t \right)^{1/2}$$ Then we have $$\int_a^b|f(x)|^2\mathrm{d}x \le \int_a^b(x-a)\int_a^x|f'(t)|^2\mathrm{d}t\mathrm{d}x \le \int_a^b |f'(t)|^2\mathrm{d}t\int_a^b(x-a)\mathrm{d}x$$ It's a few steps from here i guess