Proving the inequality $ 1+\mathbb{S}<(1+a_1) \dot\ \dots \dot\ (1+a_n)<\frac{1}{1-\mathbb{S}} $ 
Given $a_1,a_2,\dots,a_n>0$ and $\mathbb{S}=a_1+\dots+a_n<1$ Prove: $$ 1+\mathbb{S}<(1+a_1) \dot\ \dots \dot\ (1+a_n)<\frac{1}{1-\mathbb{S}}   $$

I proved the left inequality, but I can't figure the right one.
Any help appreciated.
 A: Let $S=\sum_{k=1}^n a_k<1$, where $a_k>0$ for all $k$.  
METHODOLOGY $1$:  Use the AM-GM Inequality
Using the AM-GM Inequality revelas
$$\begin{align}
\prod_{k=1}^n(1+a_k)&\le \left(\frac{\sum_{k=1}^n(1+a_k))}{n}\right)^n\\\\
&=\left(1+\frac{S}{n}\right)^n\tag 1\\\\
\end{align}$$
Now, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's inequality that (i) the sequence $\left(1+\frac xn\right)^n$ monotonically increases to $e^x$ and (ii) that the exponential function satisfies the inequality $e^x\le \frac{1}{1-x}$ for $x<1$. 
Therefore, we see that 
$$\begin{align}
\prod_{k=1}^n(1+a_k)&\le e^S\\\\
&\le \frac{1}{1-S}
\end{align}$$
as was to be shown!

METHODOLOGY $2$:  Use a Standard Inequality for the Loagarithm Function
Again, in THIS ANSWER, I showed that the logarithm satisfies the inequality $\log(x)\le x-1$.  
Hence, we have
$$\begin{align}
\log\left(\prod_{k=1}^n(1+a_k)\right)&=\sum_{k=1}^n\log(1+a_k)\\\\
&\le \sum_{k=1}^na_k\\\\
&=S
\end{align}$$
Therefore, we see that 
$$\prod_{k=1}^n(1+a_k)\le e^S\le \frac{1}{1-S}$$
as expected!
A: As you proved the left one yourself, here is the proof of the right ineq. by induction:
$$(1+a_1)(1-a_1)=1-a_1^2<1 \Rightarrow 1+a_1<\frac{1}{1-a_1}$$
Suppose for some $n-1\in \Bbb N$, $$(1+a_1)...(1+a_{n-1})<\frac{1}{1-\Bbb S_{n-1}}$$
Then if $a_1,...,a_n>0$ and $\Bbb S<1$, we have
$$(1+a_1)...(1+a_n)<\frac{1+a_n}{1-\Bbb S_{n-1}}=\frac{1+a_n}{1-\Bbb S_{n-1}}\frac{1-a_n}{1-a_n}=\frac{1-a_n^2}{1-\Bbb S_n +a_n\Bbb S_{n-1}}<\frac{1}{1-\Bbb S_n}$$
where $\Bbb S_{j}=\sum_{i=1}^{j}a_i$
Result follows
