Stuck on End of Proving Bonferroni Inequality An exercise asks me to prove "Bonferroni Inequality" and states it in the form of $$P\left( {A \cap B} \right) \ge 1 - P\left( {{A^C}} \right) - P\left( {{B^C}} \right)$$
I've searched a lot of old questions related to this and all of them seem to do some crazy math abstraction and reasoning (for my level, of course), so I just tried to do it with the "tools" I have myself.
I did the following:
$$\eqalign{
  & P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)  \cr 
  & P\left( {A \cup B} \right) - P\left( A \right) = P\left( B \right) - P\left( {A \cap B} \right)  \cr 
  & P\left( {{A^C}} \right) = P\left( B \right) - P\left( {A \cap B} \right) \cr} $$Adding $-1$ to each side yields
$$\eqalign{
  & P\left( {{A^C}} \right) - 1 = P\left( B \right) - 1 - P\left( {A \cap B} \right)  \cr 
  & P\left( {{A^C}} \right) - 1 =  - P\left( {{B^C}} \right) - P\left( {A \cap B} \right)  \cr 
  & P\left( {A \cap B} \right) = 1 - P\left( {{A^C}} \right) -  - P\left( {{B^C}} \right) \cr} $$
I don't know how to go from there ($=$) to the inequality I need. I'm assuming it's somethings obvious that I missed but I don't know how to bring it in.
On a side note, the question also asks me to generalize it for more than two sets... Is it doable following this path I chose?
 A: Not as an answer, but in order to offer an easier way. The proof of inequality is quite simple: 
$$\mathbb P(A\cap B) = 1-\mathbb P(A^c\cup B^c) \geq 1-(\mathbb P(A^c)+\mathbb P(B^c))$$
The first equality is due to $(A\cap B)^c=A^c\cup B^c$, and the second inequality is since $$\mathbb P(X\cup Y)=\mathbb P(X)+\mathbb P(Y) - \mathbb P(X\cap Y)\leq \mathbb P(X)+\mathbb P(Y).$$ 
A: FIRST APPROACH
The given inequality is equivalent to
\begin{align*}
\textbf{P}(A\cap B) \geq 1 - \textbf{P}(A^{c}) - \textbf{P}(B^{c}) \Longleftrightarrow \textbf{P}(A^{c}) + \textbf{P}(B^{c}) \geq 1 - \textbf{P}(A\cap B) = \textbf{P}(A^{c}\cup B^{c})
\end{align*}
Thus the problem is reduced to prove that the following inequality holds
\begin{align*}
\textbf{P}(A) + \textbf{P}(B) \geq \textbf{P}(A\cup B)
\end{align*}
which is true indeed. In fact, we have
\begin{align*}
\textbf{P}(A) + \textbf{P}(B) \geq \textbf{P}(A) + \textbf{P}(B) - \textbf{P}(A\cap B) = \textbf{P}(A\cup B)
\end{align*}
SECOND APPROACH
The inequality about to be proven is known as the Fréchet-Hoeffding inequality. It is stated as follows
\begin{align*}
\textbf{P}(A\cap B) \geq \max\{\textbf{P}(A) + \textbf{P}(B) - 1,0\}
\end{align*}
Indeed, we have
\begin{align*}
\textbf{P}(A\cup B) = \textbf{P}(A) + \textbf{P}(B) - \textbf{P}(A\cap B) \leq 1 \Longleftrightarrow \textbf{P}(A\cap B) \geq \textbf{P}(A) + \textbf{P}(B) - 1
\end{align*}
Since $\textbf{P}(A\cap B)\geq 0$, the claimed result holds. Based on it, we get
\begin{align*}
\textbf{P}(A\cap B) \geq \textbf{P}(A) + \textbf{P}(B) - 1 = (1 - \textbf{P}(A^{c})) + (1 - \textbf{P}(B^{c})) - 1 = 1 - \textbf{P}(A^{c}) - \textbf{P}(B^{c})
\end{align*}
and the proposed inequality holds.
GENERALIZATION FOR MORE THAN TWO SETS
First of all, lets us consider the generalized version of the inequality:
\begin{align*}
\textbf{P}(A_{1}\cap A_{2}\cap\ldots\cap A_{n}) \geq 1 - \textbf{P}(A^{c}_{1}) - \ldots - \textbf{P}(A^{c}_{n})
\end{align*}
This inequality is equivalent to
\begin{align*}
\textbf{P}(A^{c}_{1}) + \textbf{P}(A^{c}_{2}) + \ldots + \textbf{P}(A^{c}_{n}) \geq 1 - \textbf{P}(A_{1}\cap A_{2}\cap\ldots\cap A_{n}) = \textbf{P}(A^{c}_{1}\cup A^{c}_{2}\cup\ldots A^{c}_{n})
\end{align*}
that is the same as
\begin{align*}
\textbf{P}(A_{1}) + \textbf{P}(A_{2}) + \ldots + \textbf{P}(A_{n}) \geq \textbf{P}(A_{1}\cup A_{2}\cup \ldots\cup A_{n})
\end{align*}
which can be proven using the principle of mathematical induction.
