a bound about a collection of positive probability events Let $A_1,\dots,A_n$ be some events such that $\forall\ i\in[n]:P(A_i)>0$.
Prove that 
$$\frac{\sum\limits_{i=1}^n\sum\limits_{j=1}^nP(A_i\cap A_j)}{\Bigg(\sum\limits_{i=1}^nP(A_i)\Bigg)^2}\geq 1$$
I tried proving it using induction, but I somehow feel that the bound is a tight bound so it's not good to eliminate elements, since we don't know the relation between  $P(A_i)P(A_j)$ and $P(A_i\cap A_j)$.
Using induction:
base: $n=1$, we have:
$$\frac{P(A_1)}{(P(A_1))^2}\geq1\iff P(A_1)\geq P(A_1)^2$$
which is true, since $P(A_1)\in(0,1]$.
Suppose it is true for $n$ events, lets prove it for $n+1$ events. Let $A_1,\dots,A_{n
Lets say $c_n=\sum\limits_{i=1}^n\sum\limits_{j=1}^nP(A_i\cap A_j)$ and $c_{n+1}=\sum\limits_{i=1}^{n+1}\sum\limits_{j=1}^{n+1}P(A_i\cap A_j)$.
Observe that $c_{n+1} = c_n+2\sum\limits_{i=1}^nP(A_{n+1}\cap A_i)+P(A_{n+1})$.
Lets say $d_n=\sum\limits_{i=1}^nP(A_i)$ and $d_{n+1}=\sum\limits_{i=1}^{n+1}P(A_i)$.
Observe that $d_{n+1}^2 = d_n^2+2d_nP(A_{n+1})+P(A_{n+1})^2$.
So, from induction hypothesis we have $c_n\geq d_n^2$.
We need to prove that $c_{n+1}\geq d_{n+1}^2$, i.e.
$$c_n+2\sum\limits_{i=1}^nP(A_{n+1}\cap A_i)+P(A_{n+1})\geq d_n^2+2d_nP(A_{n+1})+P(A_{n+1})^2$$
I is enough to prove that
$$2\sum\limits_{i=1}^nP(A_{n+1}\cap A_i)+P(A_{n+1})\geq 2d_nP(A_{n+1})+P(A_{n+1})^2$$
And I am stuck since if we eliminate to get 
$$\sum\limits_{i=1}^nP(A_{n+1}\cap A_i)\geq d_nP(A_{n+1})$$
I find it hard, maybe it is even wrong...
 A: I forget all about co-variance...
The proof is quite simple using it:
We need to prove that 
$$\frac{\sum\limits_{i=1}^n\sum\limits_{j=1}^nP(A_i\cap A_j)}{\Bigg(\sum\limits_{i=1}^nP(A_i)\Bigg)^2}\geq 1\iff$$
$$\sum\limits_{i=1}^n\sum\limits_{j=1}^nP(A_i\cap A_j)\geq \Bigg(\sum\limits_{i=1}^nP(A_i)\Bigg)^2\iff$$
$$\sum\limits_{i=1}^n\sum\limits_{j=1}^nP(A_i\cap A_j)\geq \sum\limits_{i=1}^n\sum\limits_{j=1}^nP(A_i)(A_j)\iff$$
$$\sum\limits_{i=1}^n\sum\limits_{j=1}^nP(A_i\cap A_j)-\sum\limits_{i=1}^n\sum\limits_{j=1}^nP(A_i)(A_j)\geq0\iff$$
$$\sum\limits_{i=1}^n\sum\limits_{j=1}^n[P(A_i\cap A_j)-P(A_i)(A_j)]\geq0\iff$$
$$\sum\limits_{i=1}^n\sum\limits_{j=1}^nCOV(\mathbf{1}_{A_i},\mathbf{1}_{A_j})\geq0$$
Using bi-linearity and symmetry we get:
$$\sum\limits_{i=1}^n\sum\limits_{j=1}^nCOV(\mathbf{1}_{A_i},\mathbf{1}_{A_j})=COV(\sum\limits_{i=1}^n\mathbf{1}_{A_i},\sum\limits_{j=1}^n\mathbf{1}_{A_j})=VAR(\sum\limits_{i=1}^n\mathbf{1}_{A_i})\geq0$$ 
A: First consider the situation where all $A_i$ are mutually exclusive.
$$r=\frac{\sum\limits_{i=1}^n\sum\limits_{j=1}^nP(A_i\cap A_j)}{\Bigg(\sum\limits_{i=1}^nP(A_i)\Bigg)^2}=\frac{\sum\limits_{i=1}^nP(A_i)}{\sum\limits_{i=1}^nP(A_i)\sum\limits_{j=1}^nP(A_j)}=\frac{1}{\sum\limits_{j=1}^nP(A_j)}\ge1$$
since $s=\sum\limits_{j=1}^nP(A_j)\le1$.
Now suppose that the only non-zero intersection is between $A_1$ and $A_2$ such that $P(A_1\cap A_2)=p_{12}$. Then,
$$r=\frac{2p_{12}+\sum\limits_{i=1}^nP(A_i)}{\sum\limits_{i=1}^nP(A_i)\sum\limits_{j=1}^nP(A_j)}\ge\frac{2p_{12}+s}{(1+p_{12})s}=\frac{s+2p_{12}}{s+sp_{12}}\ge1$$
since $s=\sum\limits_{j=1}^nP(A_j)\le1+p_{12}$ and $s\le2$. I suggest continuing by including more intersections in this manner.
Since the comments request the work be shown for the next step, let's include $P(A_1\cap A_3)=p_{13}$, $P(A_2\cap A_3)=p_{23}$, and $P(A_1\cap A_2\cap A_3)=p_{123}$. 
Also let $u=P(A_1\cup A_2\cup...\cup A_n)$ so that
$s=u+p_{12}+p_{13}+p_{23}-p_{123}$. 
Then
$$r=\frac{2p_{12}+2p_{13}+2p_{23}+\sum\limits_{i=1}^nP(A_i)}{\sum\limits_{i=1}^nP(A_i)\sum\limits_{j=1}^nP(A_j)}=\frac{u+3(p_{12}+p_{13}+p_{23})-p_{123}}{\left(u+p_{12}+p_{13}+p_{23}-p_{123}\right)^2}$$
In this case indeed, it is not easy to show $r\ge1$. Perhaps looking at all of the exclusive sets that make up the union, $q_1=P(A_1\cap(A_2\cup A_3)^c)$, $q_{12}=P((A_1\cup A_2)\cap(A_3)^c)$ and $q_r=P(A_4\cup...\cup A_n)$ so that
$$u=q_1+q_2+q_3+q_r+q_{12}+q_{13}+q_{23}+q_{123}$$ then
$$r=\frac{q_1+q_2+q_3+q_r+4(q_{12}+q_{13}+q_{23})+9q_{123}}{(q_1+q_2+q_3+q_r+2(q_{12}+q_{13}+q_{23})+3q_{123})^2}
=\frac{a+4b+9c}{(a+2b+3c)^2}$$
where the only constraints are that the constants $(a,b,c)\ge0$ and $0<a+b+c\le1$. Even with this form it is not simple to see that $r\ge1$.
Consider the case that $u=a+b+c=1$ then,
$$r=\frac{1+3b+8c}{(1+b+2c)^2}=\frac{1+2b+4c+(b+4c)}{1+2b+4c+(b+2c)^2}\ge1$$
since
$$(b+c)+3c\ge(b+c)^2+2(b+c)c+c^2=(b+c)^2+(2(b+c)+c)c$$
In the case that $u=a+b+c=1-d$ it works out similarly, just messier. Is it enough to show that this holds for three intersections, to prove that it holds for the general case? The problem only deals with pair-wise intersections. I suspect there is a more elegant proof.
