Suppose we have two sets of discrete events, $A$ and $B$. Then I think it is true that:


$$2\sum\_{i \in A, j \in B}\Pr[i\ \textrm{AND}\ j] \leq \sum\_{i \in A}\Pr[i]+ \sum\_{j \in B}\Pr[j] +\sum\_{i, j \in A}\Pr[i\ \textrm{AND}\ j] + \sum\_{i, j \in B}\Pr[i\ \textrm{AND}\ j]$$


My intuition for why this should be true is just by analogy to the simple Cauchy-Schwarz like inequality for $a, b \in \mathbb{R}:$
$$2ab \leq a + b + a(a-1) + b(b-1)$$
To see why this identity is true, note:
$$a + b + a(a-1) + b(b-1) = a^2 + b^2 = 2ab + (a-b)^2 \geq 2ab$$


Is my inequality true, and is there a simple proof?



Probability space analogue of Cauchy-Schwarz inequality
yes, if $\sum\_{i,j\in A}$ means double summation (each pair $i,j$ is taken twice)
then denote $f\_i$ and $g\_i$ characteristic functions of your events from $A$ and $B$ respectively, LHS equals $2 \sum \int f\_i g\_j=\int 2(\sum f\_i)(\sum g\_j)$, RHS equals $\sum \int f\_i^2+\sum \int g\_j^2+2\sum \int f\_i f\_j+2\sum \int g\_i g\_j=\int (\sum f\_i)^2+(\sum g\_j)^2$, and RHS-LHS equals $\int (\sum f\_i-\sum g\_j)^2\geq 0$.


