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Let $p_i$ be the $i^{th}$ prime number. It seems as though the following inequality is true for all positive integers $m$ and real numbers $x>1$: $$1+\frac{1}{p_m^x}\leq\prod_{i=m+1}^{\infty}\left(1+\frac{1}{p_i^x}\right).$$ Is this correct, and, if so, is there a way to prove it? Thank you.

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The inequality does not always hold. Take $m=1$ for example, in which case the inequality is equivalent to $$ \bigg( 1+\frac1{2^x} \bigg)^2 - 1 \le \prod_p \bigg( 1+\frac1{p^x} \bigg) - 1 = \frac{\zeta(x)}{\zeta(2x)} - 1, $$ where $\zeta$ is the Riemann zeta function. However, the left-hand side is asymptotic to $2^{1-x}$ as $x\to\infty$, while the right-hand side is asymptotic to $2^{-x}$; therefore the inequality fails for large $x$. (Indeed, a calculation indicates that it fails for all $x>1.865$ or so.)

A similar argument should show that the inequality fails for every $m$, when $x$ is large enough in terms of $m$.

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