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This question already has an answer here:

Prove:$$0<a_k\in \mathbb R\quad and\quad\prod_{k=1}^n a_k =1,\quad then\quad \prod_{k=1}^n (1+a_k) \ge 2^n$$

(*) I guess that the minimum of $\prod\limits_{k=1}^n (1+a_k)$ happens when all $a_k$'s are $1$, but can't prove it. Thanks.

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marked as duplicate by Martin Sleziak, Martin R, Community Dec 31 '16 at 16:19

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it is simple $AM-GM$, we have $$1+a_1\geq 2\sqrt{a_1}$$ $$1+a_2\geq 2\sqrt{a_2}$$ .......................... $$1+a_n\geq 2\sqrt{a_n}$$ multiplying all together we get $$\prod_{k=1}^{n} 1+a_k\geq 2^n\sqrt{a_1a_2a_3...a_n}=2^n$$

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  • $\begingroup$ Equality holds if and only if $\sqrt{a_k}=1$, i.e. $a_k=1$ for all $k\in\{1,2,\ldots,n\}$. $\endgroup$ – user236182 Dec 31 '16 at 15:26

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