Suppose $f$ is twice differentiable and satisfies $f(0)=0$. Prove the inequality. $$\int_0^1 |f(x)f'(x)| dx \le\ \frac{1}{2} \int_0^1 |f'(x)|^2 dx $$
This is a problem from undergraduate math competition in Korea.
I may use the Cauchy-Schwarz inequality, but it is difficult for me to rearrange this inequality to make form for the Cauchy-Schwarz inequality.
How to prove?
$$\int f f' dx=\int f df=1/2f^2$$ Thus $$\int_0^1f(x)f'(x)dx=f(x)|_0^1=1/2f(1)^2$$ since $f(0)=0$. Let $g(x)=1$ then by Cauchy's theorem $$\left|\int (f')^2 dx\right|\left|\int g^2 dx\right|\geq\left(\int f'g\,dx\right)^2=(f)^2$$ Thus since $\int_0^1 g(x)dx=1$ $$1/2\int_0^1 (f')^2 dx\geq1/2f(1)^2=\int_0^1 f f' dx$$ The only part I did not get was the absolute value sign around $ff'$, could this be another typo?