# Prove $0<a_k\in \mathbb R$ and $\prod\limits_{k=1}^n a_k =1$, then $\prod\limits_{k=1}^n (1+a_k) \ge 2^n$ [duplicate]

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.

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

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 $$\prod_{k=1}^{n} 1+a_k\geq 2^n\sqrt{a_1a_2a_3...a_n}=2^n$$
• 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