# 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.

• Split it up by degree. The inequality is (almost) obvious then. Commented Nov 7, 2020 at 1:46

I think you guys are trying to overkill these problems. Clearly $${\rm LHS} \le \left(1 + \frac{S}{n}\right)^n = \sum_{k \ge 0}^n \frac{n!}{(n-k)!\cdot n^k} \frac{S^k}{k!} \le {\rm RHS}$$

• actually LHS $\geq\left(1+\dfrac{S}{n}\right)^n$ due to Holder. Commented May 28, 2021 at 19:34
• @dezdichado Are you sure about that? that inequality just simply came from Am-Gm Commented May 28, 2021 at 19:36
• yeah sorry i don't know what i was thinking lol Commented May 28, 2021 at 19:37

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.

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

$$1 + x_{i+1} + S + x_{i+1}S + \frac{S^2}{2!}S + x_{i+1}\frac{S^2}{2!} + \ldots + \frac{S^k}{k!} + x_{i+1} \frac{ S^k}{k!} \\ \leq 1 + S' + \frac{ S'^2}{2!} + \ldots + \frac{ S' ^{k+1} } { (k+1)!}$$