TITLE: Inequality of Product and Sum
QUESTION [6 upvotes]: Let $\forall i \in \{1,\dots,K\}, a_i \in (0,1)$ and $\sum_{i=1}^{K} a_i = 1$. Let $P_i = \prod_{j=1}^{i} (1-a_j)$ and $S_i = \sum_{j=1}^{i} a_j$. 
Is there proof or counterexample to $\forall i, (1-P_i)/S_i \geq (1-P_K)$?

REPLY [2 votes]: One can in fact show that
$$
 \frac{1-P_1}{S_1} > \frac{1-P_2}{S_2} > \ldots > \frac{1-P_K}{S_K} \, .
$$
if all $a_i \in (0, 1)$. If, in addition, $\sum_{i=1}^{K} a_i = 1$ then the last term is equal to $1-P_K$, and the desired conclusion follows.
Proof: For $1 \le i \le K-1$
$$
\begin{align}
 \frac{1-P_i}{S_i} - \frac{1-P_{i+1}}{S_{i+1}} &= \frac{(1-P_i)S_{i+1} - (1-P_{i+1})S_i}{S_i S_{i+1}} \\
&=  \frac{S_{i+1}-S_i - P_i S_{i+1} + P_{i+1} S_i}{S_i S_{i+1}} \\
&= \frac{a_{i+1} - P_i(S_i+a_{i+1}) + P_i(1-a_{i+1})S_i}{S_i S_{i+1}} \\
&= \frac{a_{i+1} (1-P_i(1+S_i))}{S_i S_{i+1}}
\end{align}
$$
and that is positive because
$$
\begin{align}
 P_i(1+S_i)& \le 
 (1-a_1)\cdots(1-a_i)(1+a_1)\cdots(1+a_i) \\
 &= (1-a_1^2)\cdots(1-a_i^2) \\
 &< 1 \, .
\end{align}
$$