Before I begin, I saw this inequality on another site without any info on where it comes from, etc. It might not even be true, but I certainly don't know either way. I am simply asking if anyone either has seen something similar or actually has an idea of how to prove/disprove it. Here is the inequality:
Let $f\in\mathcal{C}^3([0, 1])$ with $|f'''|\le 1$, $f(1)=f(0)=f'(0)=0$. Prove that $$\sqrt{3}|f(x)|\le x(1-x)\sqrt{\int_0^x \frac{|f(t)|}{t(1-t)}}$$
A few things: the equality is achieved by $\frac{x^2(1-x)}{6}$. If we define $g(x) = \frac{6f(x)}{x(1-x)}$, then it is easy to prove that $x\ge |g(x)|$, and the inequality becomes $$\frac{g(x)^2}{2}\le \int_0^x |g| \Leftrightarrow \left(\sqrt{\int_0^x |g|}\right)' \le \frac{1}{\sqrt{2}}$$ but now the third derivative condition is a mess.
I feel I have seen a similar type of functional inequalities in the past, but I am not really sure. Any ideas?