I'm trying to prove the following:
If $A_1,A_2,\cdots, A_n $ are independent events, then the probability that non of the $ A_i$ occurs is less than $e^{-\sum^{n}_{i=1}P(A_i)} $.
I was exploring different alternatives. The first was writing $P\left(\left( \bigcup A_i \right)^c \right)$ as the probability of the intersection of the complements, but I'm not sure if this will works because I cannot use the hypothesis of independence since the $A_i^c$ are not independent.
Then, looking for an option in which the probability of the intersection shows, I was trying to prove that $ P\left(\left( \bigcup^{n}_{i=1}A_{i} \right) \right) \geq 1-e^{-\sum^{n} P(A_i)}$, which allows me to use induction and the independence hypothesis. To my regret, I wasn't able to get to any place with this idea too.
I have been trying but now I feel stuck. Any ideas?
Let $\Pr(A_i)=p_i$ for all $i$.
Hint:
Use De Morgan's law to see that $\Pr\left(\bigcup_i A_i\right)=1-\Pr\left(\bigcap A_i^c\right)$
As the $A_i$'s are independent, so are the $A_i^c$'s. This means $\Pr\left(\bigcap A_i^c\right)=\prod_i(1-p_i)$.
Take logarithm on both sides of the last equation and expand the $\ln$ series in the r.h.s to prove that
$$\ln \left(\Pr\left(\bigcap A_i^c\right)\right)\leqslant-\sum_ip_i$$