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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3By Hölder’s inequality $$\prod_{k=1}^n (1+a_k) \ge \left(\left(\sqrt[n]{1}\right)^n+\prod_{k=1}^n \sqrt[n]{a_k}\right)^n=2^n$$ – user236182 Dec 31 '16 at 15:20
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@user236182 Nice find. One more tool in my tool box... – Simply Beautiful Art Dec 31 '16 at 15:21
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3$$ (1+a_k)\geq 2\sqrt{a_k} $$ is enough. – Jack D'Aurizio Dec 31 '16 at 15:21
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See also What is the minimum value of $(1 + a_1)(1 + a_2). . .(1 + a_n)$? and other posts linked there. Found using Approach0. – Martin Sleziak Dec 31 '16 at 15:32
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Follows also from Prove $(x+r_1) \cdots (x+r_n) \geq (x+(r_1 \cdots r_n)^{1/n})^{n}$. – Martin R Dec 31 '16 at 15:48
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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 $$\prod_{k=1}^{n} 1+a_k\geq 2^n\sqrt{a_1a_2a_3...a_n}=2^n$$
Dr. Sonnhard Graubner
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Equality holds if and only if $\sqrt{a_k}=1$, i.e. $a_k=1$ for all $k\in{1,2,\ldots,n}$. – user236182 Dec 31 '16 at 15:26