Inequality of integral of $f, f^2, xf, xf^2$ Let $f$ be an differentiable increasing nonnegative function from $[0,1]$ to $\mathbb{R}$ which satisfy:


*

*$\forall x\in [0,1]:f(x)\ge0$

*$\forall x\in [0,1]:f'(x)\ge0$

*$\sup_{x\in[0,1]}f(x)=M$

*$\sup_{x\in[0,1]}f'(x)=m$
Prove that
$$\bigg(\int_0^1x\big(f(x)\big)^2dx\bigg)\cdot\bigg(\int_0^1f(x)dx\bigg)-\bigg(\int_0^1\big(f(x)\big)^2dx\bigg)\cdot\bigg(\int_0^1xf(x)dx\bigg)\le\frac{M^2m}{12}$$
The funny part is that I have proved that $LHS\le0.087M^2m$, but I'm never able to reduce the bound... Annoying...
 A: Let us use Fubini to write the left hand side as
$$\int_0^1 \int_0^1 f(x)f(y) x (f(x)-f(y)) dx dy\tag{1}$$
Now note that by the mean value theorem, for every $x,y\in (0,1)$, there is a $\xi\in(0,1)$ (depending on $x,y$) such that
$$f^\prime(\xi)(x-y)=f(x)-f(y)$$
Thus, $(1)$ becomes
$$\int_0^1 \int_0^1 f(x) f(y) f^\prime(\xi) x(x-y) dx dy\tag{2}$$
Let us split $(2)$ into two parts:
$$\int_0^1 \int_y^1 f(x) f(y) f^\prime(\xi) x(x-y) dx dy-\int_0^1 \int_0^y f(x) f(y) f^\prime(\xi) x(y-x) dx dy\tag{3}$$
Using Fubini and renaming the variables $x$ and $y$, the second summand can be rewritten as:
$$\begin{align}
\int_0^1 \int_0^y f(x) f(y) f^\prime(\xi) x(y-x) dx dy &=\int_0^1 \int_x^1 f(x) f(y) f^\prime(\xi) x(y-x) dy dx\\
&= \int_0^1 \int_y^1 f(y) f(x) f^\prime(\xi) y(x-y)dxdy
\end{align}
$$
Note here that we used that the dependence of $\xi$ on $x,y$ is symmetric.
Plugging this into $(3)$ we get that the LHS equals
$$ \int_0^1 \int_y^1 f(x) f(y) f^\prime(\xi) (x-y)^2 dx dy$$
which we can now estimate by
$$M^2 m \int_0^1\int_y^1 (x-y)^2 dx dy=\frac{M^2 m}{12}$$
