TITLE: Prove that $\int_0^1xf(x) dx \leq \frac{2}{3}(\int_0^1f(x)dx)^2$
QUESTION [8 upvotes]: Let $f:[0,1] \to [0,+\infty)$ be non negative continuous concave function, such that $f(0) = 1$. Prove that $$\int_0^1xf(x) dx \leq \frac{2}{3}(\int_0^1f(x)dx)^2$$
Work: My professor left us some hints, but I am not sure if I am using them correctly, also I am not sure how to use all of them. Hints are: First hint, substitute $t=\lambda x$ in integral $$F(x)=\int_0^xf(t)dt.$$ What I have got is $$F(x)=\lambda \int_0^xf(\lambda x)dt.$$ Second hint is, using concavity and $f(0)=1$ prove that $$F(x)\geq \frac{x}{2}(1+f(x)).$$ Okay so, for concave functions there is inequality such as this $$f(y)\leq f(x)-f'(x)(y-x).$$ But I haven't got anything from that. Last hint is, using partial integration, conclude that $$\int_0^1 x f(x) dx \leq F(1) - \frac{1}{4} - \frac{1}{2} \int_0^1 x f(x) dx.$$ I think I have proven that, here is what I did: From partial integration I have got $$\int_0^1 x f(x) dx = x F(x){|}_0^1 - \int_0^1 F(x) dx \leq F(1) - \int_0^1 \frac{x}{2} (1+f(x)) dx \leq F(1) - \frac{1}{4} - \frac{1}{2} \int_0^1xf(x) dx$$ From that we have $$\int_0^1x f(x) dx \leq \frac{2}{3} (F(1) - \frac{1}{4}).$$ Using basic inequality $A^2-A+1/4 \geq 0$ we get $F(1) - \frac{1}{4} \leq F^2(1)$ and proof is finished.
Conclusion: My question is how to prove the second hint and where we use substitution here, is that step necessary?
Edit: I have realised that substitution I wrote above was not correct, should be $$\int_0^1 f(\lambda)d\lambda.$$ I guess we use it to make integral be $\int_0^1$. Still haven't proven second hint. Any help will be welcome.

REPLY [8 votes]: By concavity, $f(\lambda x + (1 - \lambda)0) \ge \lambda f(x) + (1 - \lambda)f(0) = \lambda f(x) + (1 - \lambda)$, so
\begin{align}
F(x) &= \int_0^x f(t) \text d t = x \int_0^1 f(\lambda x) \text d\lambda
\\
&\ge x \int_0 ^1 (\lambda f(x) + (1 - \lambda)) \text d \lambda
\\
&= x \left( \frac 12 f(x) + \frac 12 \right).
\end{align}
Now, using this along with integration by parts with $u = x$ and $\text d v = f(x)$ yields
\begin{align}
\int_0 ^1 x f(x) \text d x &= x F(x)|_0^1 - \int_0^1 F(x) \text d x 
\\
&\le F(1) - \frac 12\int_0^1 x \left(f(x) + 1\right) \text d x
\\
&= F(1) - \frac 14- \frac 12\int_0^1 x f(x)  \text d x,
\end{align}
At which point algebra reveals
$$
\int_0^1 x f(x) \text dx \le \frac 23 \left(F(1) - \frac 14\right). 
$$
Since $0 \le (F(1) - 1/2)^2 = F(1)^2 - F(1) + 1/4$, the last inequality and the definition of $F$ give:
$$
\int_0^1 x f(x) \text d x \le \frac 23 \left(\int_0^1 f(x) \text dx \right)^2.
$$