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 ?

Probably the most succinct proof, shamelessly adapted from another answer:
If $P(B)=0$ or $P(B)=1$ the inequality is trivial, so we may assume $P(B)\in (0,1)$. Then note that $$\begin{align}|P(A\cap B)-P(A)P(B)|
&= \left|P(B)P(B^c)\left[ \frac{P(A\cap B)}{P(B)P(B^c)} - \frac{P(A\cap B)+P(A\cap B^c)}{P(B^c)} \right]\right|\\ 
&= \left|P(B)P(B^c)\left[\frac{P(A\cap B)(1-P(B))}{P(B)P(B^c)}-\frac{P(A\cap B^c)}{P(B^c)} \right]\right|\\
&=P(B)P(B^c)|P(A\mid B)-P(A\mid B^c)|\\
&=P(B)(1-P(B))\underbrace{|P(A\mid B)-P(A\mid B^c)|}_{\leq 1}\\
&\leq \frac 14
\end{align}$$
because $x(1-x)\leq \frac 14$ when $x\in [0,1]$.
