Prove if $\{B_n\}$ are events such that $\sum_{i=1}^nP(B_i)>n-1 \Rightarrow P(\bigcap_{i=1}^n B_i)>0.$

Question

In the case when $n=2$, we have that if $$P(A\cup B)>1, \text{ then }$$ $$P(A\cup B)=P(A)+P(B)-P(A\cap B)>1-P(A\cap B),$$ from here, we have that If $P(A\cap B)=0$, then $P(A\cup B)>1 !!$, so, necesarily, $P(A\cap B)>0.$

In the case when $n>1$, I think that inclusion-exclusion formula can be used in a similar way as in case when $n=1$, but I can't conclude the proof.

Answer 1

From $\sum_{i=1}^nP(B_i)>n-1$ we obtain $$\sum_{i=1}^nP(B_i^c)<1$$also since $P(\bigcup _{i=1}^nB_i^c)\le \sum_{i=1}^nP(B_i^c)$ (union bound) we obtain $$P(\bigcup _{i=1}^nB_i^c)<1$$or $$1-P(\bigcup _{i=1}^nB_i^c)>0\implies P\Bigg((\bigcup _{i=1}^nB_i^c)^c\Bigg)>0\implies P(\bigcap_{i=1}^nB_i)>0$$