Bonferonni inequality I have seen many types of bonferonni inequalities and how to solve them, but not quite this one.
$$P\left(\bigcup_{i=1}^n A_i\right)\geq \sum_{i=1}^nP(A_i) - \sum_{i<j}P(A_i\cap A_j)$$
I've started trying using induction:
Let n=2
then we know $P(A_1\cup A_2) = P(A_1) + P(A_2) - P(A_1\cap A_2)$
and we know that $P(A_1\cap A_2) \geq 0$
Therefore this is true for n=2
Assume its true for n=k then:
$$P\left(\bigcup_{i=1}^k A_i\right) \geq \sum_{i=1}^kP(A_i) - \sum_{i<j}P(A_i\cap A_j)$$
let $B=\bigcup_{i=1}^k A_i$
Then
$$P(B\cup A_{k+1}) \geq P(B) + P(A_{k+1}) - \sum_{i<k+1}P(B\cap A_{k+1})$$
$=P(\sum_{i=1}^{k+1} A_i) -\sum_{i<k+1}(B\cap A_{k+1})$
But i think in the last two steps i'm doing something wrong. Any help?

For brevity let $C=\sum_{i=1}^k P(A_i)-\sum_{1\leq i<j\leq k}P(A_i\cap A_j).$ We have $P(B)\geq C.$
We also have $P(B\cap A_{k+1})=P(\cup_{i=1}^k(A_i\cap A_{k+1}))\leq \sum_{i=1}^kP(A_i\cap A_{k+1}).$
So $\;-P(B\cap A_{k+1})\geq -\sum_{i=1}^kP(A_i\cap A_{k+1}).$ $$\text { Therefore }\quad  P(\cup_{i=1}^{k+1}A_i)=P(B\cup A_{k+1})\geq$$ $$\geq P(B)+P(A_{k+1})-P(B\cap A_{k+1})\geq$$ $$\geq C+P(A_{k+1})-P(B\cap A_{k+1})\geq$$ $$\geq C+P(A_{k+1})-\sum_{i=1}^k P(A_i\cap A_{k+1})=$$ $$=\sum_{i=1}^{k+1}P(A_i)-\sum_{1\leq i<j\leq k+1}P(A_i\cap A_j).$$
