Proving $|P(A\cap B)-P(A)P(B)|\leq \frac{1}4$ 
Let $A$ and $B$ be two events of a probability space. Prove that  $\displaystyle|P(A\cap B)-P(A)P(B)|\leq \frac{1}4$

I think it's a very challenging problem, and I've made no progress so far ...
Can someone give me a hint ?

Note that
$\begin{align}
|P(A\cap B) - P(A)P(B)| &= |E[1_A1_B] - E[1_A]E[1_B]|\\
&= |Cov(1_A,1_B)|\\
&= |E[\{1_A-P(A)\}\{1_B-P(B)\}]| \\
&\leq \sqrt{V(1_A)}\sqrt{V(1_B)}
\end{align}$
where the last inequality follows from Cauchy-Schwarz for the following inner product on $L^2(P)$
$$\langle X,Y\rangle = E[XY].$$
Next, $V(1_A) = E[1_A^2] - E[1_A]^2 = E[1_A] - E[1_A]^2 = E[1_A](1-E[1_A])\leq \frac 14$, hence the claim.
