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The inequality is as follows: $$a_1,a_2,...,a_n \in\Bbb R ; \prod_{i=1}^n a_i=1 \Rightarrow \prod_{i=1}^n (a_i+1) \ge 2^n$$

What I've tried:

I've seen that as you multiply each term you get a 1, wich means you will eventually get to a one on the left side after you multiply n terms, so the right term would turn into a $2^n-1$. I've noticed also that as you multiply the terms you get a product of all the terms you multiply, wich would mean that as you get to n terms multiplied, you would get something like $\prod_{i=1}^n a_i$ on the left side, wich is 1, according to the problem. so in the right side you would have $2^n -1 -1 = 2^n - 2$. And that's it, I don't know what to do now.

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    Are you sure about $1+i$, shouldn't it be $1+a_i$? – 2017-01-16
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    Another one: [Prove $0$\prod\limits_{k=1}^n a_k =1$, then $\prod\limits_{k=1}^n (1+a_k) \ge 2^n$](http://math.stackexchange.com/questions/2078552/prove-0a-k-in-mathbb-r-and-prod-limits-k-1n-a-k-1-then-prod-limits). – 2017-01-16

2 Answers 2

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You can use AM-GM:

$$\frac{1+a_1}{2}\ge \sqrt{a_i}\rightarrow 1+a_i \ge 2\sqrt{a_i}$$

so

$$(1+a_1)(1+a_2)...(1+a_n) \ge 2^n \sqrt{a_1a_2...a_n}=2^n$$

2

The most direct way is to use Huygens inequality

$(1+a_1)(1+a_2)\ldots(1+a_n)\ge(1+\sqrt[n]{a_1a_1\ldots a_n})^n=2^n$

Or you can just use Holder, which is a generalization of this inequality.