Inequality of Product and Sum 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 $\forall i, (1-P_i)/S_i \geq (1-P_K)$?
 A: Case $i = K$ is trivial, so we can assume $S_i < 1$.
$P_K = P_i \cdot \prod_{j=i+1}^K (1 - a_j)$ and $\sum_{j=1+1}^K a_j = 1 - S_i$, we have $P_K \geqslant P_i \cdot (1 - a_{i + 1} - a_{i + 2}) \cdot \prod_{j=i+3}^K \ldots \geqslant P_i \cdot(1 - a_{i + 1} - \ldots - a_K) = P_i \cdot S_i$ and so $1 - P_K \leqslant 1 - P_i \cdot S_i$, so it's enough to prove $\frac{1 - P_i}{S_i} \geqslant 1 - P_i \cdot S_i$.
$$\frac{1 - P_i}{S_i} \geqslant 1 - P_i \cdot S_i$$
$$1 - P_i \geqslant S_i - P_i S_i^2$$
$$1 - S_i \geqslant P_i (1 - S_i^2)$$
$$1 \geqslant P_i (1 + S_i)$$
Note that $(1 + a_1) \cdot (1 + a_2) \cdot \ldots (1 + a_i) \geqslant (1 + S_i)$, so it's enough to prove
$$1 \geqslant \prod_{j=1}^i (1 - a_j) \cdot \prod_{j=1}^i (1 + a_i)$$
$$1 \geqslant \prod_{j=1}^i (1 - a_j^2)$$
The last is true, as non-empty product of positive numbers each of which is less than $1$ is itself less than $1$.
A: 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
$$
\begin{align}
 P_i(1+S_i)& \le 
 (1-a_1)\cdots(1-a_i)(1+a_1)\cdots(1+a_i) \\
 &= (1-a_1^2)\cdots(1-a_i^2) \\
 &< 1 \, .
\end{align}
$$
