An inequality involving the sum of the probability of intersection of sets with same probability Suppose $P(A_i)=p$ for any $1\leq i\leq n.$
Prove that
$$n(n-1)p^2-\sum_{j,k;j\neq k}P(A_j\cap A_k)\leq n.$$
Hint: consider Cauchy or Jensen's inequality.

My attempt:
\begin{eqnarray*}
\text{LHS}&=&n(n-1)p^2-\sum_{j,k;j\neq k}P(A_j\cap A_k)\\
&=& \sum_{j,k;j\neq k}P(A_j)P(A_k)-P(A_j\cap A_k)\\
&=& \sum_{j,k;j\neq k}\mathbb{E}\mathbf{1}_{A_j}\mathbb{E}\mathbf{1}_{A_k}-\mathbb{E}\mathbf{1}_{A_j}\mathbf{1}_{A_k}
\end{eqnarray*}
Should I apply Cauchy here?
Any help would be appreciated. Thanks!
 A: Consider the binary random variable $\mathcal{B}_i$ that take value $1$ on $A_i$ and $0$ on its complement. Then the correlation of the random variables $\mathcal{B}_i$ and $\mathcal{B}_j$  is 
$$ c_{ij} = \frac{P(A_i \cap A_j) - P(A_i) \cdot P(A_j)}{\sqrt{p(1-p)}\cdot \sqrt{p(1-p)}}= \frac{P(A_i \cap A_j) - p^2}{p(1-p)}$$
The correlation matrix $(c_{ij})$ is positive semi-definite and therefore the sum of its elements is $\ge 0$. So we have 
$$\sum_{i,j} \frac{P(A_i \cap A_j) - p^2}{p(1-p)} \ge 0$$ or, equivalently,
$$n + \sum_{i\ne j} \frac{P(A_i \cap A_j) - p^2}{p(1-p)} \ge 0$$
Therefore
$$n(n-1)p^2 - \sum_{i\ne j} P(A_i \cap A_j) \le n p (1-p)$$ 
a better inequality than the one stated. Let's reformulate our inequality as follows:
$$\frac{\sum_{i<j } P(A_i \cap A_j)}{\binom{n}{2}}\ge p^2 - \frac{p(1-p)}{n-1}$$
Notice that if $n$ is large then the average size of the pairwise intersection is larger than something approaching $p^2$. One cannot get better than $p^2$, since there exist independent $A_i$ with size $p$. 
Observation: for a fixed $n$ the lower bound for $\sum_{i<j } P(A_i \cap A_j)$ stated above is optimal provided that $p$ is a rational number with denominator $n$ : $0, 1/n, 2/n, \ldots, 1$
and is not optimal otherwise. The lower bound as a function of $p$ is the piecewise linear interpolation of the above values at $p=0, 1/n,\ldots, 1$. 
