Prove that $P( A_1 \cap A_2 \cap \ldots A_n) \geq P(A_1) + P(A_2) + \ldots + P(A_n) - (n-1)$ Problem:

Show that
$$
P( A_1 \cap A_2 \cap \ldots A_n)
 \geq P(A_1) + P(A_2) + \ldots + P(A_n) - (n-1)
$$

Answer:
Recall that:
$$ P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
We can rewrite this as:
$$P(A \cap B) = P(A) + P(B) -  P(A \cup B) $$
I am going to prove this by induction on $n$.
case $n = 1$
We need to prove that:
$$ P(A_1) \geq P(A_1) - (1-1)$$
Since $1-1 = 1$ it is obviously true.
case $n = 2$
We need to prove that:
$$ P(A_1 \cap A_2) \geq P(A_1) + P(A_2) - (2-1)$$
We have:
\begin{align*}
P(A_1 \cap A_2) &= P(A_1) + P(A_2) -  P(A_1 \cup A_2) \\
P(A_1) + P(A_2) -  P(A_1 \cup A_2) &\geq P(A_1) + P(A_2) - 1 \\
P(A_1) + P(A_2) -  P(A_1 \cup A_2) &\geq P(A_1) + P(A_2) - (2-1) \\
\end{align*}
Now we assume it for $n = i$. This means we have:
$$ P( A_1 \cap A_2 \cap ... A_{i}) \geq P(A_1) + P(A_2) + .. + P(A_{i}) - (i-1) $$
Now we prove it for $n=i+1$. We need to prove that
$$ P( A_1 \cap A_2 \cap ... A_{i+1}) \geq P(A_1) + P(A_2) + .. + P(A_{i+1}) - (i+1-1)$$
\begin{align*}
P( A_1 \cap A_2 \cap ... A_{i+1}) &=
 P( A_1 \cap A_2 \cap ... A_{i}) + P(A_{i+1})
  - P( \left(  A_1 \cap A_2 \cap ... A_{i}\right)  \cup A_{i+1} ) \\
%P( A_1 \cap A_2 \cap ... A_{i+1}) \geq P(A_1) + P(A_2) + .. + P(A_{i+1}) - (i+1-1) \\
\end{align*}
Is my solution correct so far? How do I finish the job?

From your last line, apply $$-P((A_1 \cap \cdots \cap A_i) \cup A_{i+1}) \ge -1$$ and $$P(A_1 \cap A_2 \cap \cdots \cap A_i) \ge P(A_1) + \cdots + P(A_i) - (i-1).$$
