Probability space analogue of Cauchy-Schwarz inequality

Question

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?

Answer 1

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 $\int (\sum f_i-\sum g_j)^2\geq 0$.