# 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 $$\forall i, (1-P_i)/S_i \geq (1-P_K)$$?

• What is $k$ in $P_k$. You have tagged this question as "inequality", but you are asking about an equality. Is this a typo? – Pantelis Sopasakis May 15 '19 at 22:49
• Thanks for pointing out the mistake with $K$. Changed it to the inequality form. – Soumya Basu May 15 '19 at 22:55

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 $$1$$.
• Thanks! It is a brilliant solution. Specifically, the observation that $P_K \geq P_i S_i$ is great. – Soumya Basu May 16 '19 at 3:32
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}