Prove that for events $A$, $B$ such that: $\mathbb{P}(B) > 0$, it’s true that: $\mathbb{P}(A|A \cup B) ≥ \mathbb{P}(A|B)$

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True or False : $P( B \mid A \cup B) \geq P(B \mid A)$

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Prove that for events $A$, $B$ such that: $\mathbb{P}(B) > 0$, it's true that: $\mathbb{P}(A|A \cup B) ≥ \mathbb{P}(A|B)$.

I tried to do the following:
$$\mathbb{P}(A|A \cup B) ≥ \mathbb{P}(A|B)$$

$$\frac{\mathbb{P} (A \cap (A \cup B))}{\mathbb{P}(A \cup B)} ≥ \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$$

$$\frac{\mathbb{P}(A)}{\mathbb{P}(A) + \mathbb{P}(A \cap B) + \mathbb{P}(B)} ≥ \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$$

$$\mathbb{P}(A) \mathbb{P}(B) \geq \mathbb{P}(A)\mathbb{P}(A \cap B) + \mathbb{P}(A \cap B)\mathbb{P}(A \cap B) + \mathbb{P}(B)\mathbb{P}(A \cap B)$$

Or alternatively:

$$\mathbb{P}(A) \mathbb{P}(B) \geq \mathbb{P}(A \cap B)\mathbb{P}(A \cup B)$$

Eather way, I don't see why that would be the case

Answer 1

Let $a = \mathbb{P}(A \cap B), b = \mathbb{P}(A - B), c = \mathbb{P}(B - A)$. We have $$\mathbb{P}(A)\mathbb{P}(B) = (a + b)(a + c),$$ $$\mathbb{P}(A\cap B)\mathbb{P}(A\cup B) = a(a + b + c),$$ and $$(a + b)(a + c) - a(a + b + c) = bc \geq 0.$$