Prove that $P(A \cap B ) \geq 1 - P(\bar{A}) - P(\bar{B})$
This is what I got:
So I know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Rearranging for $P(A \cap B)$
$P(A \cap B) = P(A) + P(B) - P(A \cup B)$
Substitute $P(A) = 1-P(\bar{A})$ and $P(B) = 1-P(\bar{B})$
$P(A \cap B) = 1 - P(\bar{A}) - P(\bar{B}) + 1 - P(A \cup B)$
Substitute $P(\overline{A \cup B}) = 1 - P(A \cup B)$
$P(A \cap B) = 1 - P(\bar{A}) - P(\bar{B}) + P(\overline{A \cup B})$
So $P(A \cap B) \geq 1 - P(\bar{A}) - P(\bar{B}) \enspace\enspace\enspace \text{ Since } P(\overline{A \cup B})\leq 1$
I'm not sure if my proof is correct or not and wanted to ask for some verification and correction if i'm wrong.
Sorry if the structure is messy, I'm not that familiar with latex.
Yes! It is a valid proof, except that the last line should read "$\text{since} \ P(\overline{A \cup B}) \geq 0$" because $A = B + C$ implies $A \geq B$ only if $C \geq 0$.
However, your proof is more complex than it needs to be. We can get there faster by starting from a version of one of the terms you introduce later on:
$$P(A \cap B) = 1 - P(\overline{A \cap B}).$$
We can use the underlying Boolean logic to take a shortcut. By De Morgan's law,
$$\overline{A \cap B} = \bar{A} \cup \bar{B},$$
and therefore
$$P(\overline{A \cap B}) = P(\bar{A} \cup \bar{B}).$$
We also know that
$$P(\bar{A} \cup \bar{B}) = P(\bar{A}) + P(\bar{B}) - P(\bar{A} \cap \bar{B}).$$
All that remains is to chain these together, negating where necessary:
$$P(A \cap B) = 1 - P(\bar{A}) - P(\bar{B}) + P(\bar{A} \cap \bar{B}).$$
Since the last term above is a probability, we know it must be greater than or equal to zero, and the desired result immediately follows.