If $x_1 + x_2 + \ldots + x_n \leq \frac{1}{3}$, then $ (1 - x_1) (1 - x_2) \cdots (1 - x_n) \geq \frac{2}{3}$.

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I'm trying to solve the below problem from a University of Oxford problem sheet.

Given $n$ positive integers $x_1, x_2, \ldots, x_n$ such that $x_1 + x_2 + \ldots + x_n \leq \frac{1}{3}$, prove by induction that $$ (1 - x_1) (1 - x_2) \cdots (1 - x_n) \geq \frac{2}{3}. $$

I'm not completely sure whether the way this is written implies that $n$ needs to be at least $2$. Otherwise, I'd be inclined to write $x_1, \ldots, x_n$ to emphasize that $n = 1$ is possible. Regardless, here's my attempt. The problem gives a hint to consider $x_1, \ldots, x_{n-1}, x_n + x_{n+1}$ for the inductive step. I'm not sure how to use this or where the sum of the $n$th and $(n+1)$st terms comes from. That seems to me to sacrifice generality.

Let $n = 1$. Then $x_1 \leq \frac{1}{3}$, so $1 - x_1 \geq 1 - \frac{1}{3} = \frac{2}{3}$. Suppose that this statement holds for $k$, and suppose that $x_1, \ldots, x_k + x_{k+1} \leq \frac{1}{3}$. As the $x_i$ are positive, we know that $$ \sum\limits_{i=1}^{k} x_i \leq \sum\limits_{i=1}^{k+1} x_i \leq \frac{1}{3}. $$ The induction hypothesis then implies that $$ \prod\limits_{i=1}^k (1 - x_i) \geq \frac{2}{3}. $$

I'm stuck at this point. I'm not sure whether I should write out an $n = 2$ base case to make the transition from $k$ to $k+1$. I think I need to show that $1 - x_{k+1} > 0$. That means that $x_{k+1} < 1$. If the sum $\sum\limits_{i=1}^k x_i \leq \frac{1}{3}$ and $x_{k+1} \geq 1$, then $\sum\limits_{i=1}^{k+1} x_i > \frac{1}{3}$, which is a contradiction. Is that on the right track?

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For the inductive step, we want to show that if $x_1,\dots,x_k,x_{k+1}$ are positive numbers such that $x_1+\dots+x_k+x_{k+1}\le 1/3$, then $(1-x_1)\cdot\dots\cdot(1-x_{k+1})\ge 2/3$.

What we know by inductive hypothesis is that for any positive numbers $y_1,\dots,y_k$ such that $y_1+\dots+y_k\le 1/3$, we have $(1-y_1)\cdot\dots\cdot(1-y_k)\ge2/3$. The hint is suggesting you consider the particular assignment $$ y_1 = x_1,\dots,y_k=x_k+x_{k+1}. $$ When you apply the inductive hypothesis to these particular $y_1,\dots,y_k$, what do you get?

Note that it doesn't have to be just this assignment which would work. Another assignment which would work is $$ y_1 = x_1 + x_2, y_2 = x_3, \dots, y_k = x_{k+1}. $$ The point is to make an assignment that connects the expression $x_1+\dots+x_{k+1}$ with $y_1+\dots+y_k$ for some numbers $y_1,\dots,y_k$, so that we can make use of our inductive assumption regarding a sum of $k$ positive numbers.

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Key fact here: $(1 - a)(1 - b) > 1 - a - b$ for $0 < a, b < 1$. The result you are looking for can be shown by repeatedly using this fact... so try to turn this repeated use into an induction argument. The hint you are given is a suggestion on how to formalize this idea.