TITLE: If $S = x_1 + x_2 + .. + x_n$, Prove that $ (1+x_1)(1+x_2)..(1+x_n) \le 1 + S + \frac{S^2}{2!} + .. + \frac{S^n}{n!}$
QUESTION [11 upvotes]: Let $x_1, x_2, \ldots ,x_n$ be positive real numbers, and let $$ S = x_1 + x_2 + \cdots + x_n.$$
Prove that  $$ (1+x_1)(1+x_2)\ldots(1+x_n)  \le 1 + S + \frac{S^2}{2!} + \cdots + \frac{S^n}{n!}$$
Here's my attempt:
Case 1 : Let $x_1 = x_2= x$ (When all terms are equal)
$LHS = (1+x)(1+x) = 1+2x+x^2$
$RHS = 1+\frac{x+x}{1!}+\frac{(x+x)^2}{2!}=1+2x+\color{blue}{2}x^2$
Hence when all terms are equal, $LHS<RHS$.
Case 2 : $x_1\ne x_2$
$LHS=(1+x_1)(1+x_2)=1+{x_1}{x_2}+(x_1+x_2)$
$RHS=1+ (x_1+x_2)+\frac{(x_1+x_2)^2}{2!} =1+{x_1}{x_2}+(x_1+x_2)+\color{blue}{\left(\frac{x_1^2+x_2^2}{2}\right)}$
Hence when even one term is not equal, $LHS\lt RHS$.
From the two cases, it's clear that under no circumstance $LHS \gt RHS$. And hence the (partial) proof.
Is there any way I can improve my proof? If you look again I've considered only two terms and my intuition tells me that it would apply to the entire range but how do I state it mathematically? Or can you think of a better or more rigorous proof than this?
Thanks again!

REPLY [3 votes]: I use this fact :$$(x_1+x_2+x_3...+x_n)^n=\sum_{k_1+k_2+...k_j=n} \dfrac{n!}{k_1!k_2!...}x_1^{k_1}x_2^{k_2}...x_n^{k_j}$$
$$1 \leq 1\\(x_1+x_2...+x_n)\leq S\\(x_1x_2+x_1x_3+...+x_{n-1}x_n) \leq \frac12S^2=\frac12(2(x_1x_2+x_1x_3+...)+(x_1^2+x_2^2+...x_n^2))\\(x_1 x_2x_3+x_1x_2x_4+...)\leq\frac{1}{3!}S^3=\frac{1}{6}(\color{red} {\frac{3!}{1!1!1!}(x_1x_2x_3+x_1x_2x_4+...)}+\frac{3!}{2!1!}(x_1^2x_2+x_2^2x_1+...)+\frac{3!}{3!}(x_1^3+x_2^3+...+x_n^3))\\$$
$$(x_1 x_2x_3x_4+x_1x_2x_3x_5+...)\leq\frac{1}{4!}S^4=\frac{1}{24}(\color{red} {\frac{34!}{1!1!1!1!}(x_1x_2x_3x_4+x_1x_2x_3x_5+...)}+\frac{4!}{3!1!}(x_1^3x_2+x_2^3x_1+...)+\frac{4!}{2!2!}(x_1^2x_2^2+...)+\frac{4!}{4!}(x_1^4+x_2^4+...+x_n^4))$$ and so on 
now we have 
$$1\leq1\\+(x_1+x_2+...x_n)\leq\frac{1}{2!}S^2\\
+(x_1x_2+x_1x_3+...)\leq\frac{1}{3!}S^3\\
+(x_1x_2x_3+x_1x_2x_4+...)\leq\frac{1}{4!}S^4\\+...\\
+(x_1x_2x_3...x_n)\leq\frac{1}{n!}S^n\\
1+(x_1+x_2+...x_n)+(x_1x_2+x_1x_3+...)+...(x_1x_2...x_n)\leq 1+s+\frac{S^2}{2}+...+\frac{S^n}{n!}\\$$
and 
$$(1+x_1)(1+x_2)...(1+x_n)=1+(x_1+x_2+...x_n)+(x_1x_2+x_1x_3+...)+...(x_1x_2...x_n)$$proof is complete now