How do I prove the following inequality (link included)?

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For any two events $A$ and $B$, show that $$(P(A\cap B))^2+(P(A\cap B^c))^2+(P(A^c\cap B))^2+(P(A^c\cap B^c))^2\ge\frac 1 4$$

I tried solving the question by writing $A \cap B^c = A - (A \cap B)$ and similiarly expanding the other brackets. But I am not able to proceed any forward.

I also tried applying the Boole's inequality and then squaring the respective terms, but still didn't get the required answer.

Can somebody help me with this question, please?

I would be extremely grateful.

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We want to solve

$\min a^2+b^2 + c^2 + d^2$

subject to $a+b + c+ d=1$

where $a,b,c,d \ge 0$.

Hint:

Apply Cauchy-Schwarz on $(a,b,c,d)$ and $(1,1,1,1)$.