Miklos Schweitzer 1968 P11 by Renyi Source: Miklos Schweitzer 1968, Problem 11

Let $ A_1,...,A_n$ be arbitrary events in a probability field. Denote by $ C_k$ the event that at least $ k$ of $ A_1,...,A_n$ occur. Prove that $$\prod_{k=1}^n P(C_k) \leq  \prod_{k=1}^n P(A_k).$$

Attempt: I believe the first step to tackle this would be to employ Bayes rule:
$$P(C_n)=P(A_1\cap\cdots\cap A_n)=P(A_2\cap\cdots\cap A_n|A_1)P(A_1)$$
as here we accrue a $P(A_1)$ term times something less than or equal to one which is promising for the inequality to show. Then my guess would be to continue to establish similar correspondences with the other $P(A_i)$ terms. I try that by further employing Bayes to get
$$P(C_n)=P(A_1)P(A_2|A_1\cap A_3\cap\cdots\cap A_n)P(A_3\cap\cdots\cap A_n|A_1).$$
However this seems to lead to a dead end (at least with $C_n$) because I'm conditioning the other $A_k$ terms on some measurable subset of the sample space, so I can't obtain the other $P(A_k)$ terms unconditioned without magic. Another approach or follow-up that seems messy is to use sub-additivity of $P$ to look at the $C_k$ terms for $k<n$:
$$P(C_k)=P\left(\bigcup\limits_{i_1\neq i_2\neq\cdots\neq i_k} A_{i_1}\cap\cdots\cap A_{i_k}\right)\leq \sum\limits_{i_1\neq i_2\neq\cdots\neq i_k} P\left(A_{i_1}\cap\cdots\cap A_{i_k}\right)$$
but um, aside from applying Bayes here to get $P(A_k)$ times some constant $\leq 1$, this doesn't help as we are adding these factors, not multiplying.
Question: I would appreciate tactics/hints here (preferably related to my line of reasoning if possible), not full blown solutions - thanks!
 A: I think I solved it. First do the case $n = 2$, which can be done directly using $P(A \cap B) = P(A) - P(A \setminus B)$, $P(A \cup B) = P(B) + P(A \setminus B)$. Then notice that if $A_1 \supset \dots \supset A_n$, then $C_k = A_k$ and we have equality. The strategy to reduce to this case is by modifying the sets until we have $A_1 \supset \dots \supset A_n$. The $n = 2$ case will help you do the modifications.
A: Write $X_i = \mathbf{1}_{A_i}$ and let $S_n = X_1 + \cdots + X_n$. Then we have $C_k = \{ S_n \geq k \}$, and so, the problem boils down to proving
$$ \prod_{k=1}^{n} \mathbf{P}(S_n \geq k) \leq \prod_{k=1}^{n} \mathbf{P}(X_k = 1). \tag{1} $$
Now we prove this by induction. Note that we always have
$$ \mathbf{P}(A)\mathbf{P}(B) \geq \mathbf{P}(A\cup B)\mathbf{P}(A \cap B). \tag{*} $$
Then
Base case. When $n = 1$, the claim is trivial.
Induction step. Suppose $\text{(1)}$ holds for some $n > 1$. Then
\begin{align*}
\prod_{k=1}^{n+1} \mathbf{P}(X_k = 1)
&\geq \mathbf{P}(X_{n+1} = 1)\left[ \prod_{k=1}^{n} \mathbf{P}(S_n \geq k) \right].
\end{align*}
Also, note that
\begin{align*}
&\mathbf{P}(X_{n+1} = 1, S_{n} \geq k-1)\mathbf{P}(S_n \geq k) \\[0.25em]
&\geq \mathbf{P}\left(\begin{gathered} (X_{n+1} = 1, S_{n} \geq k-1) \\ \text{ or } (S_{n} \geq k) \end{gathered}\right) \cdot \mathbf{P}\left( \begin{gathered} (X_{n+1} = 1, S_{n} \geq k-1) \\[0.25em] \text{ and } (S_{n} \geq k) \end{gathered}\right) \\
&= \mathbf{P}(S_{n+1} \geq k) \mathbf{P}(X_{n+1} = 1, S_{n} \geq k).
\end{align*}
Now use this relation for $k = 1, \ldots, n$ to establish the induction step.
