Proving Inequality Involving Sums and Products of Variables in a Probabilistic Analysis

Question

I'm exploring a probabilistic analysis problem where I have variables (or, probabilities) $x_1,\ldots, x_n \in (0,1)$ in the range $(0,1)$ satisfying $\sum^n_{i=1} x_i \in (0,1)$. I aim to prove the inequality: $$\frac{1-\prod^{k+1}_{i = 1} (1-x_i)}{\sum^{k+1}_{i=1} x_i} \leq \frac{1-\prod^{k}_{i = 1} (1-x_i)}{\sum^k_{i=1} x_i}.$$ For all $k \in \{1,2,\ldots,n-1\}$. This inequality essentially suggests that augmenting the denominator by adding $x_{k+1}$ to the denominator while multiplying the second term in the nominator by $(1-x_{k+1})$ overall can only decrease the overall expression.

I've attempted to employ the arithmetic/geometric mean inequality, but I haven't succeeded so far. Can anyone provide insights or alternative approaches that could help in proving this inequality?

Answer 1

For ease of notation / laziness, let $ S_K = \sum_{i=1}^{K} x_i, P_K = \prod_{i=1}^K (1 - x_i) , $.

• Cross multiplying, WTS $ S_k - S_k P_{k+1} \leq S_{k+1} - S_{k+1} P_k $
• Shifting and grouping terms, WTS $x_{k+1} P_k + S_kP_k [ 1 - ( 1-x_{k+1}) ] \leq x_{k+1}$.
• Dividing by $x_{k+1}$, WTS $P_k ( 1 + S_k) \leq 1$. -> OP stated in comments that they arrived at this step.
• This follows from $ \frac{1}{1 -x_i} \geq 1 + x_i $ for $ x_ i \in (0,1 )$.

$$\frac{1}{ P_k} \geq \prod ( 1 + x_i ) \geq 1 + \sum x_i + \sum_{i\neq j} x_i x_j + \ldots \geq 1 + \sum x_i.$$