Inequality $\prod_{k=1}^n (1+a_k) \ge (n+1)^n \prod_{k=1}^n a_k.$ For non-negative $a_k$, $k =1,2,\dots,n$ such that $a_1+a_2+\dots+a_n=1$, then
$$\prod_{k=1}^n (1+a_k) \ge (n+1)^n \prod_{k=1}^n a_k.$$
I used induction but it became complicated. A detailed answer will be appreciated.
 A: Inspired by Real numbers inequality on AoPS: Let $S = \sum_{k=1}^n a_k$ and $P = \prod_{k=1}^n a_k$. Then, using the AM-GM inequality,
$$
 \prod_{i=1}^n(1+a_i) = \prod_{i=1}^n(S+a_i) \ge \prod_{i=1}^n \bigl( (n+1)\sqrt[n+1]{P a_i}\bigr) = (n+1)^n P \, .
$$
A: Let $ n $ be a positive integer. We have : $$ \prod_{k=1}^{n}{\left(1+\frac{1}{a_{k}}\right)}=\sum_{k=0}^{n}{\sum_{1\leq i_{1}<\cdots <i_{k}\leq n}{\frac{1}{a_{i_{1}}\cdots a_{i_{k}}}}} $$
Given some $ k\in\mathbb{N} $, using AM-GM inequality, we have : $$ \frac{1}{\binom{n}{k}}\sum_{1\leq i_{1}<\cdots <i_{k}\leq n}{\frac{1}{a_{i_{1}}\cdots a_{i_{k}}}}\geq \left(\prod_{1\leq i_{1}<\cdots <i_{k}\leq n}{\frac{1}{a_{i_{1}}\cdots a_{i_{k}}}}\right)^{\frac{1}{\binom{n}{k}}}=\left(\prod_{k=1}^{n}{\frac{1}{a_{k}}}\right)^{\frac{\binom{n-1}{k-1}}{\binom{n}{k}}}=\left(\prod_{k=1}^{n}{\frac{1}{a_{k}}}\right)^{\frac{k}{n}} $$
Using AM-GM again, we have $ \left(\prod\limits_{k=1}^{n}{a_{k}}\right)^{\frac{1}{n}}\leq\frac{1}{n}\sum\limits_{k=1}^{n}{a_{k}}=\frac{1}{n} $, and hence : $$ \left(\prod_{k=1}^{n}{\frac{1}{a_{k}}}\right)^{\frac{1}{n}}\geq n $$
Thus : $$ \sum_{1\leq i_{1}<\cdots <i_{k}\leq n}{\frac{1}{a_{i_{1}}\cdots a_{i_{k}}}}\geq\binom{n}{k}n^{k} $$
Hence : $$ \prod_{k=1}^{n}{\left(1+\frac{1}{a_{k}}\right)}\geq\sum_{k=0}^{n}{\binom{n}{k}n^{k}}=\left(n+1\right)^{n} $$
A: This is the solution if $0<a_k<1$.
LHS is $P_1 = e^{\log P_1} = e^{ \sum_{k=1}^{n} \log (1+a_k)} \geq e^{\sum_{k=1}^{n} (a_k - \frac{a_k^2}{2})} = e^{\frac{1}{2}}$
This inequality is due to Taylor series expansion for $0<x:$
$$
x-\frac{x^2}{2}< \log(1+x) <x
$$
and the fact that $0<a_k^2 < a_k <1$.
RHS can be rewritten as 
$$
(n+1)^n \prod_k a_k = e^{\sum_{k=1}^{n}\log ((n+1)^n a_k)} = e^{\sum_{k=1}^{n}\log b_k} < 1
$$
where $b_k = (n+1) a_k$, and, since $0<a_k<1, \log b_k <0$ 
Putting these all together, RHS $<$ LHS
A: For a simple answer only using $\text{AM}\geq \text{GM}$:  
consider dividing each side by $\big(\prod_{k=1}^n a_k\big)$  and re-writing the desired inequality as
$\prod_{k=1}^n \big(1+\frac{1}{a_k}\big) \geq \big(n+1\big)^n$
now take nth roots to see this is equivalent to  
$\Big(\prod_{k=1}^n \big(1+\frac{1}{a_k}\big)\Big)^\frac{1}{n} $
$\geq \big(\prod_{k=1}^n\frac{1}{a_k}\big)^\frac{1}{n} +\big(\prod_{k=1}^n 1\big)^\frac{1}{n} $
$ \geq \frac{1}{\frac{1}{n}}+1$
$=n +1$ 
where the first inequality comes from superadditivity of the geometric mean (which may be proven with a one-liner using $\text{AM}\geq \text{GM}$ ) and the second inequality comes from observing $\frac{1}{n}\geq \big(\prod_{k=1}^n a_k\big)^\frac{1}{n}$ which holds from $\text{AM}\geq \text{GM}$, and then inverting.    
The original inequality is met with equality iff $\frac{1}{n} = a_1 = a_2=...=a_n$ 
