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.
I can show that for $x &gt; 0$ and $r_{i} &gt; 0$ we have $$ \left(\, x + r_{1}\,\right)\ldots\left(\, x + r_{n}\,\right)\ \geq\ \left[\, x + \left(\, r_{1}\ldots r_{n}\,\right)^{1/n}\,\right]^{n}.$$ However, I can't do this using straight up induction, strong or weak. Can someone do this?   You can show the inequality holds for $n$ if it holds for $n+1$ using $r_{n+1} = (r_1 \cdots r_n)^{1/n}$, so going down isn't a problem, but going up has eluded me. Additionally, you can show that if it holds for $n=a$ and $n=b$ then it holds for $n=ab$. Using powers of two and the above is sort of a way to prove the above by induction but it certainly isn't "normal".  You can also prove the above using Lagrange multipliers (which isn't surprising). The last proof I know is where you compare coefficients of $x^k$ and use the AM-GM inequality.
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