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Let $x_1\in(0,1)$ and $a_1,\ldots,a_n\ge-1$ reals. We know that

\begin{equation} \prod_{i=1}^n (1+x_1a_i) < 1 \end{equation}

Does it then also hold true that

\begin{equation} \prod_{i=1}^n (1+x_2a_i) < 1 \end{equation}

where $x_2$ is some other real in $(0,1)$?

This is somewhat clear for $n=1$, but I'm not sure about $n$ in general.

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up vote 2 down vote accepted

Yes: $\displaystyle f(x) = \prod_{i=1}^n (1 + x a_i)$ is a continuous function of $x$, so if $f(x_1) < 1$ for some $x_1 \in (0,1)$, there is some $\delta > 0$ such that $f(x) < 1$ for all $x \in (x_1-\delta,x_1+\delta)$.

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No. Make $x_1$ quite small and the $a_i$ medium sized (e.g., all $> .5$) so the first inequality holds. Then choose $x_2$ large enough so that all the $1+x_2 a_i$ exceed $1$.

This may be a misinterpretation of the question, but it is how I understand it.

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