Given a probability space $(Ω, \mathfrak{A}, P)$ and events $A_1,A_2 . . . , A_n ∈ \mathfrak{A}.$ Show that $$P\bigg(\bigcup_{j=1}^{n} A\bigg) \leq \min_{1\leq i\leq n} \bigg( \sum_{j=1}^{n}P(A_j)- \sum_{j=1,j\neq i}^{n} P(A_j\cap A_i)\bigg).$$
I first tried to show a similar result : For $i\in\{1,...,n \}$: $$P\bigg(\bigcup_{j=1}^{n} A_j\bigg) \leq P(A_i)+ \sum_{j=1,j\neq i}^{n} P(A_j\cap A_i^c). $$
Utilizing the inclusion-exclusion principle: $$P(A_i)+\sum_{j=1,J\neq i} P(A_j \cap A_i^c)= P(A_i)+\sum_{j=1,J\neq i} P(A_j\setminus A_i)$$ Beyond this I couldn't move forward really. I'd appreciate the help.
There's no inclusion-exclusion needed. Just note that $$P(A_j) - P(A_j \cap A_i) = P(A_j \cap \bar{A_i})$$ and $$A_i \cup \bigcup_{j \neq i} (A_j \cap \bar{A_i}) = \bigcup_{j} A_j.$$ Thus the result follows by the union bound.