This should be a simple application of Maclaurin.

Just write LS as: 1 + (x_1 +x_2+...+x_n) + (x_1*x_2 + x1*x_3 + ... +x_(n-1)*x_n) + ... + (x_1*x_2*...x_n). Then compare each term a_i on LS with the term b_i on RS, the terms are counted from the left. It's very easy to verify that a_i =< b_i for every i. This gets the required result.

@Karth how we can conclude that obviously $n!\leq n^{n-k}k!$?

Using $k+1,...,n-1,n $ do not exceed$n$.

shobber wrote: Let $x_1$, $x_2$, $\cdots$, $x_n$ 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 solution $ (1+x_1)(1+x_2)...(1+x_n)\leq (1+\frac{S}{n})^n= \sum_{i = 0}^{n}\binom{n}{i}\frac{S^{n-i}}{n^{n-i}} $ $\binom{n}{k}\frac{S^{n-k}}{n^{n-k}}\le \frac{S^{n-k}}{(n-k)!}$ $ n!\leq k!n^{n-k} $

Feridimo wrote: shobber wrote: Let $x_1$, $x_2$, $\cdots$, $x_n$ 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 solution $ (1+x_1)(1+x_2)...(1+x_n)\leq (1+\frac{S}{n})^n=\sum_{i = 0}^{n}\binom{n}{i}\frac{S^{n-i}}{n^{n-i}} $ $\binom{n}{k}\frac{S^{n-k}}{n^{n-k}}\le \frac{S^{n-k}}{(n-k)!}$ $ n!\leq k!n^{n-k} $ $ n!\leq k!n^{n-k} $ $ Proof $ $ k+1\leq n $ $ k+2\leq n $ . . . $ n-1\leq n$ $ n=n $ $ (k+1)(k+2)...(n-1)n\leq n*n*...*n=n^{n-k} $ $ \frac{n!}{k!}\leq n^{n-k} $ $ n!\leq k!n^{n-k} $

$$ (1+x_1)(1+x_2)...(1+x_n)\leq (1+\frac{S}{n})^n= \sum_{i = 0}^{n}\binom{n}{i}\frac{S^{n-i}}{n^{n-i}}\leq \sum_{i = 0}^{n}\frac{S^{n-k}}{(n-k)!}\implies n!\leq k!n^{n-k} $$$$ (k+1)(k+2)...(n-1)n\leq n*n*...*n=n^{n-k}\implies \frac{n!}{k!}\leq n^{n-k} \implies n!\leq k!n^{n-k} $$

When $n=1$, we have the equality case. When $n\geq 2$, the LHS expanded are the $i^{\textrm{th}}$ elementary symmetric polynomials, $e_i$ added together where $0\leq i \leq n$ and for $i=0,1$, it is equal to $1, S$ on the RHS. For all $2\leq i \leq n-1$, we can show that the $i^{\textrm{th}}$ elementary symmetric polynomial multiplied by $i!$ is included in $S^i$ which is obvious since $S^i$ contains all the ways that terms of $e_i$ can be formed because for each term in $e_i$, it can be formed in $i!$ ways in $S^i$. Then, for $i=n,$ it is suffices to show that $n!x_1x_2\ldots x_n\leq S^n$ which follows from the fact that $n!\leq n^n$ for $n\geq 1. \blacksquare$

Solution by induction. Fix positive reals $x_1,x_2,\ldots, x_n$. We proceed by induction and base case is trivial, therefore it is left as an exercise to the reader. Assume this is true for some $k\geq 1$: \[ \prod_{i=1}^{k}(1 + x_1) \leq \sum_{i=0}^{k} \frac{S_k^i}{i!} \]We want to show that this is true also for $k+1$: \[ \prod_{i=1}^{k+1}(1 + x_1) \leq \sum_{i=0}^{k+1} \frac{S_{k+1}^i}{i!} \]Here, $S_k=\sum_{i=1}^{k} x_i$ and $S_{k+1}=\sum_{i=1}^{k+1} x_i$. Using inductive hypothesis, we want to show that $(1+x_{k+1})\left(\sum_{i=0}^{k} \frac{S_k^i}{i!}\right)\leq \sum_{i=0}^{k+1} \frac{S_{k+1}^i}{i!}$. We claim that $\frac{S_k^{i+1}}{(i+1)!}+x_{k+1}\frac{S_k^i}{i!}\leq \frac{S_{k+1}^{i+1}}{(i+1)!}$ for every $k-1>i>1$, from which the inequality right above follows. $$\frac{S_k^{i+1}}{(i+1)!}+x_{k+1}\frac{S_k^i}{i!}\leq \frac{S_{k+1}^{i+1}}{(i+1)!}\Longleftrightarrow S_k^{i+1}+(i+1)x_{k+1}S_k^i\leq S_{k+1}^{i+1}\Longleftrightarrow (i+1)x_{k+1}S_k^i\leq x_{k+1}\left(\sum_{j=0}^{i} S_{k}^{i-j}S_{k+1}^{j}\right),$$which is true, since $S_{k}\leq S_{k+1}$. Therefore, induction is complete, meaning inducting up to any $n$, we get desired inequality.

(a)Key claim is that \[\frac{S^k}{k!} \geq x_1x_2\cdots x_k+x_1x_2\cdots x_{k-1}x_{k+1}+\cdots +x_{n+1-k}\cdots x_n\]where the RHS is the $\binom{n}{k}$ products of $k$ distinct terms. (b)Notice that upon expansion, $S^k$ will have $k!$ copies of each of these terms, since not only that $S^k$ has any product of $k$ distinct terms, their $k!$ internal permutations will also be present. (c) Using the lemma, we add the terms together and we are done.

$\color{red} \boxed{\textbf{SOLUTION}}$ $$\prod_{i=1}^{n}(1 + x_1) \leq (\frac{n+S}{n})^n= (1+\frac{S}{n})^n =1+S+\frac{(n-1)S^2}{n\times 2!}+\frac{(n-1)(n-2)S^3}{n^2\times 3!}+...\frac{(n-1)(n-2)...1\times S^n}{n^{n-1} \times n!} \leq 1+S+\frac{nS^2}{n\times 2!}+\frac{n^{2}S^3}{n^2\times 3!}+...\frac{n^{n-1}\times S^n}{n^{n-1} \times n!} = 1 + S + \frac{S^2}{2!} + \frac{S^3}{3!}+...+ \frac{S^n}{n!} \blacksquare$$

By, AM-GM, it follows, $(1+x_1)(1+x_2)(1+x_3)......(1+x_n) \le (\frac{S+n}{n})^n=(1+\frac{S}{n})^n.$ Now, this is $1+S+\frac{S^2(n-1)}{n\cdot 2!}+\frac{S^3(n-1)(n-2)}{n^2\cdot 3!}+....+\frac{S^n(n-1)!}{n^{n-1}\cdot n!} \le 1+S+\frac{S^2\cdot n}{n \cdot 2!}+.......+\frac{n^{n-1}\cdot S^n}{n^{n-1} \cdot n!},$ which is the desired.

Expand the LHS. For each $0\le i\le n$ we get $$\sum_{\text{sym}} x_1\dots x_i \le \frac{\binom{n}{i}}{n^i}S^i = \frac{n(n-1)\dots (n-i+1)}{n^i}\cdot\frac{S^i}{i!}\le \frac{S^i}{i!}$$ and summing up yields the desired.

AM-GM gives \[\prod_{i} (1+x_i) \le \left(\frac{S+n}{n} \right)^n = \left(1+\frac{S}{n} \right)^n.\] Expand the right hand side to get \[\left(1+\frac{S}{n} \right)^n = \sum_{i=0}^n \binom{n}{i} \frac{S^i}{n^i} = \sum_{i=0}^n \frac{n(n-1)(n-2)\dots(n-i+1)}{n^i} \frac{S^i}{i!} \le \sum_{i=0}^n \frac{S^i}{i!}. \ \square\]