Strengthening an inequality of exponential series

Question

A question posits that, show: $$(1+x_1)(1+x_2)\ldots(1+x_n)\\ \leq 1+\dfrac{S}{1!}+\dfrac{S^2}{2!}+\ldots\dfrac{S^n}{n!}$$ Where $S=\sum x_i$, $x_i\in\mathbb{R}$

Now, the RHS looked like the exponential series so I naturally thought to use the inequality that: $$e^x\geq1+x$$

However, this proves the inequality for the infinite series, whenever it converges. How can we prove this for just upto n terms?

Answer 1

The statement holds for non-negative $x_i$, otherwise it is false: see the comments. So the proof uses the necessary extra condition. In fact, we may assume that all the $x_i$ are positive, as zeros do not make a difference in either of the two sides.

Let $0\leq k\leq n$. We denote by $m_k$ the sum of all degree $k$ monomials obtained by expanding the product $(1+x_1)\cdots (1+x_n)$. Given $n\geq i_1\geq i_2\geq \cdots \geq i_n\geq 0$ such that $i_1+\cdots +i_n=k$, the average of the monomials $x_{\sigma(1)}^{i_1}x_{\sigma(2)}^{i_2} \cdots x_{\sigma(n)}^{i_n}$ on all $n!$ permutations $\sigma$ is less than the special average of the sum of monomials for the sequence $i_1=1, i_2=1, \ldots, i_k=1, i_{k+1}=0, \ldots, i_n=0$. This is Muirhead's inequality (see https://en.wikipedia.org/wiki/Muirhead%27s_inequality), provided that all the $x_i$ are positive. The special average is $\frac{m_k}{\binom{n}{k}}$.

Hence, by expanding $S^k= (x_1+\cdots +x_n)^k$ we obtain $n^k$ monomials which in average are at least $\frac{m_k}{\binom{n}{k}}$ (as it is the weighted average of some averages that are all greater than or equal to $\frac{m_k}{\binom{n}{k}}$). So $\frac{S^k}{n^k}\geq \frac{m_k}{\binom{n}{k}}\geq \frac{m_k}{n^k/k!}$, and consequently, $m_k\leq \frac{S^k}{k!}$.

Your problem follows easily by adding up these estimations for all $0\leq k\leq n$:

$$(1+x_1)\cdots (1+x_n)= \sum\limits_{k=0}^{n} m_k\leq \sum\limits_{k=0}^{n} \frac{S^k}{n^k}$$

Answer 2

It's true for non-negative variables.

Indeed, for $n\geq k\geq l\geq1$ by Maclaurin we obtain: $$\left(\frac{\sum\limits_{1\leq i_1<i_2<...<i_k\leq n}x_{i_1}x_{i_2}...x_{i_k}}{\binom{n}{k}}\right)^{\frac{1}{k}}\leq\left(\frac{\sum\limits_{1\leq i_1<i_2<...<i_l\leq n}x_{i_1}x_{i_2}...x_{i_l}}{\binom{n}{l}}\right)^{\frac{1}{l}},$$ which gives $$\prod_{i=1}^n(1+x_i)=1+S+\sum_{1\leq i<j\leq n}x_ix_j+...+\prod_{i=1}^nx_i\leq$$ $$\leq1+S+\frac{\binom{n}{2}S^2}{n^2}+\frac{\binom{n}{3}S^3}{n^3}+...+\frac{S^n}{n^n}\leq$$ $$\leq1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+...+\frac{S^n}{n!}.$$ I used that $$\frac{\binom{n}{k}}{n^k}=\frac{n(n-1)...(n-k+1)}{n^kk!}\leq\frac{1}{k!}.$$

Answer 3

I propose my solution as well. Noticing that:

$f(x): x \rightarrow log(1+x)$

is convex on $[0,+\infty]$, we have, applying convexity inequality on $(x_1,...,x_n)$ with equal weights:

$\prod_i (1+x_i) \le (1+\frac{S}{n})^n$

Now it is easy to see that:

$(1+\frac{S}{n})^n=\sum_{i=0}^n {n \choose i} \frac{S^i}{n^i}\le 1+...+S^n/n!$.

just by comparing the coefficients of the monomials term by term. Combining the previous two inequalities we have an alternative derivation with respect to the ones proposed, maybe more elementary.