Prove that $\prod_{i=1}^n(1+x_i)\leq \sum_{i=0}^n\frac{S^i}{i!}$, where $x_i\in\mathbb{R^+}$. 
Let $x_1$, $x_2$, $\ldots$, $x_n$ be positive real numbers, and let $$S=x_1+x_2+\cdots+x_n.$$ Prove that $$(1+x_1)(1+x_2)\cdots(1+x_n)\leq 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots+\frac{S^n}{n!}.$$

My first thought is about using induction. For $n=1$, $LHS=1+x_1\leq1+S=RHS$.
Now I suppose that this inequality holds true for $n=k$; that is, $$(1+x_1)(1+x_2)\cdots(1+x_k)\leq 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots +\frac{S^k}{k!}.$$
And here I am stuck. I can't really see how I can find the relation between when $n=k$ and $n=k+1$.
Perhaps induction may not be the way?  Any hints or suggestions will be much appreciated. Thanks.
 A: I think you guys are trying to overkill these problems. Clearly
$${\rm LHS} \le \left(1 + \frac{S}{n}\right)^n = \sum_{k \ge 0}^n \frac{n!}{(n-k)!\cdot n^k} \frac{S^k}{k!} \le {\rm RHS}$$
A: Show that
$$ i! \sum_{1 \leq a_1 < a_2 \ldots < a_i \leq n } x_{a_1} x_{a_2} \ldots x_{a_i}  \leq S^i.$$
This is almost immediately obvious by expanding the RHS and noticing that we have a ton of extra terms.
In fact, we have strict inequality when $n \geq 2$.
Then, sum from $i = 0$ to $n$ to obtain the desired inequality.
A: Induction, as you requested, works too. Just do it. Where are you stuck?

$$(1+x_1)(1+x_2)\cdots(1+x_k)\leq 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots +\frac{S^k}{k!}.$$
Let $ S' = \sum_{i=1}^{k+1} x_i = S + x_{k+1}.$
$\prod_{i=1}^{k+1} (1 + x_i) \leq (1+x_{i+1}) (1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots +\frac{S^k}{k!}) \\
= 1 + x_{i+1} + S + x_{i+1}S + \frac{S^2}{2!}S + x_{i+1}\frac{S^2}{2!} + \ldots + \frac{S^k}{k!} + x_{i+1} \frac{ S^k}{k!}     $
Claim $ S^i ( S + (i+1) x_{i+1} ) \leq S' ^{i+1}$.
This is true because the sum of the terms on both sides are equal to $(i+1)S'$, and the terms are evenly distributed on the RHS.
Corollary $x_{i+1} \frac{S^i}{i!} + \frac{S^{i+1}}{(i+1)!} \leq \frac{S'^{i+1}}{{i+1}!}$
Now, do a sum of
$ 1 \leq 1 $
$x_{i+1} \frac{S^i}{i!} + \frac{S^{i+1}}{(i+1)!} \leq \frac{S'^{i+1}}{{i+1}!}$ for $ i = 1$ to $k$
$x_{i+1} \frac{ S^k}{k!} \leq \frac{ S'^{k+1} } { (k+1)!}$.
to conclude that
$1 + x_{i+1} + S + x_{i+1}S + \frac{S^2}{2!}S + x_{i+1}\frac{S^2}{2!} + \ldots + \frac{S^k}{k!} + x_{i+1} \frac{ S^k}{k!}    \\
\leq 1 + S' + \frac{ S'^2}{2!} + \ldots + \frac{ S' ^{k+1} } { (k+1)!} $
