Prove that $P\left(A_{1}+A_{2}+\ldots+A_{n}\right) \leq P\left(A_{1}\right)+P\left(A_{2}\right)+\ldots+P\left(A_{n}\right)$ Question:  For $n$ events $A_{1}, A_{2}, \ldots . ., A_{n}$ in a probability space show that
$$
P\left(A_{1} A_{2} \ldots A_{n}\right) \geq P\left(A_{1}\right)+P\left(A_{2}\right)+\ldots+P\left(A_{n}\right)-(n-1)
$$
Hence deduce $P\left(A_{1}+A_{2}+\ldots+A_{n}\right) \leq P\left(A_{1}\right)+P\left(A_{2}\right)+\ldots+P\left(A_{n}\right)$.
Progress: Proof :
We have by Boole's inequality
$$
P\left(\bar{A}_{1}+\bar{A}_{2}+\cdots+\bar{A}_{n}\right) \leqslant P\left(\bar{A}_{1}\right)+P\left(\bar{A}_{2}\right)++\cdots+P\left(\bar{A}_{n}\right)
$$
or. $\quad P \overline{\left(A_{1} A_{2} \ldots \ldots A_{n}\right)} \leqslant P\left(\bar{A}_{1}\right)+P\left(\bar{A}_{2}\right)+\ldots+P\left(\bar{A}_{n}\right)$
by De Morgan's law.
$\begin{array}{ll}\text { or, } & 1-P\left(A_{1} A_{2} \ldots \ldots A_{n}\right) \leqslant P\left(\bar{A}_{1}\right)+P\left(\bar{A}_{2}\right)+\cdots+P\left(\bar{A}_{n}\right) \\ \text { or, } & P\left(A_{1} A_{2} \ldots \ldots A_{n}\right) \geqslant 1-\sum_{i=1}^{n} P\left(\bar{A}_{i}\right)\end{array}$.
Now $P\left(\bar{A}_{i}\right)=1-P\left(A_{i}\right)\implies 1-\sum_{i=1}^{n} P\left(\bar{A}_{i}\right)=1-\sum_{i=1}^{n}\left\{1-P\left(A_{i}\right)\right\}=\sum_{i=1}^{n} P\left(A_{i}\right)-(n-1)$.
From these we can prove the first part.
How can I deduce the last part from first part?

Hint $:$ Apply the first part on $P\left (A_1^c A_2^c \cdots A_n^c \right )$ and use De Morgan's law to deduce the second part.
