union intersection I've been looking into set theory and came across this
$$|A| + |B| \geq |A\cap B| + |A\cup B|
$$
By looking at this and drawing it out I feel like the statement holds. could someone confirm this and if possible show me a proof as to why it holds?
 A: You even have the equality : $$|A \cup B| = |A| + |B| - |A \cap B|$$
A: In your example in comments it seems to me you have gotten the numbers wrong. The sum of the sizes is $|A|+|B|=4+3=7$. The size of the intersection is $|A\cap B|=|\{3,4\}|=2$. The size of the union is $|A\cup B|=|\{1,2,3,4,5\}|=5$. So we have $4+3=7=2+5$.
Indeed, equality holds in any case. stity's answer should give you a hand in proving that. Once you accept that equality, the equality I state you should try to prove instead of your inequality should become obvious.
A: For finite sets:
Basic common sense:  If X and Y are disjoint (i.e. X $\cap$ Y = $\emptyset$).  Than |X $\cup$ Y| = |X| + |Y|.  This should be obvious.  X has |X| elements and Y has |Y| different elements and X $\cup$ Y is precisely the the set containing all the elements of X and all the elements of why.
Let |A| = a, |B| = b, and |A $\cap$ B| = m.  Let A - (A $\cap$ B) = {a $\in$ A| a $\notin$ A $\cap$ B}.  A, and  A - (A $\cap$ B) are disjoint and ( A - (A $\cap$ B)) $\cup$ (A $\cap$ B).  So  A - (A $\cap$ B) = a - m.
Likewise  B - (A $\cap$ B) = b -m.
Then A $\cup$ B = ( A - (A $\cap$ B)) $\cup$ (A $\cap$ B) $\cup$ (B - (A $\cap$ B)).  This is a union of 3 disjoint sets so
|A $\cup$ B| = (a - m) + (m) + (b -m) = a + b -m = |A| + |B| - |A $\cap$ B| and
So |A| + |B| =|A $\cup$ B| + |A $\cap$ B|.
