Proving $(P(AB))^2+(P(AB^c))^2+(P(A^cB))^2+(P(A^cB^c))^2\ge\frac{1}{4}$ for any events $A$ and $B$

Question

Show that for any two events $A$ and $B$, $$(P(AB))^2+(P(AB^c))^2+(P(A^cB))^2+(P(A^cB^c))^2\ge\frac{1}{4}$$

Using the Inclusion-Exclusion formula, De Morgan's law and $P(AB^c)=P(A)-P(AB)$, I was able to expand the $\rm{l.h.s.}$ as $$(x+y)(x+y-4z)+2z(2z+1)+x(x-2)+y(y-2)+1\quad...\quad(*),$$ where $x=P(A),y=P(B)$ and $z=P(AB)$.

However I'm not being able to group the terms involved in the $\rm{l.h.s.}$ to form a 'suitable' expression and probably the way I have grouped them is not correct.

The expression $(*)$ is subject to the condition

$\max\{0,x+y-1\}\le z\le \min\{x,y\}$ and obviously $0\le x,y,z\le 1$, but I am unable to use this.

Is there an elegant/slick way to show the inequality? It probably can be shown without a lot of work and it's possible I'm missing something trivial here.

Answer 1

With Cauchy-Schwarz inequality, you get that \begin{align}\sum_{i=1}^41^2\sum_{i=1}^4p_i^2\ge \left(\sum_{i=1}^4p_i\cdot1\right)^2&=\left(P(AB)+P(AB^c)+P(A^cB)+P(A^cB^c)\right)^2\\[0.2cm]&=\left(P(A)+P(A^c)\right)^2=1\end{align}