How to prove $\prod_{k=1}^n (1+a_k)\leq \sum_{j=0}^n\left(\sum_{k=1}^n a_k\right)^j$

Question

If $a_{i} \geq 0$ , then $(1+a_{1})(1+a_{2})\cdots (1+a_{n}) \leq 1+(a_{1}+a_{2}+\cdots+a_{n})+(a_{1}+a_{2}+\cdots+a_{n})^{2}+\cdots+(a_{1}+a_{2}+\cdots+a_{n})^{n}$

I found it in the book Theory and Applications of Infinite Series. I was not capable of proving it. And I love understanding everything while I am studying.

I am studying infinite products from this book.

Answer 1

We transform the left-hand side of (1) \begin{align*} \prod_{k=1}^n (1+a_k)\leq \sum_{j=0}^n\left(\sum_{k=1}^n a_k\right)^j\tag{1} \end{align*} until the inequality becomes obvious. In the following we use the notation $[n]:=\{1,2,\ldots,n\}$.

We obtain \begin{align*} \color{blue}{\prod_{k=1}^n\left(1+a_k\right)} &=\sum_{S\subseteq [n]}\prod_{k\in S} a_k\tag{2}\\ &=\sum_{j=0}^n\sum_{{S\subseteq [n]}\atop{|S|=j}}\prod_{k\in S}a_k\tag{3}\\ &=\sum_{j=0}^n\sum_{{l_1+l_2+\cdots+l_n=j}\atop{l_k\in\{0,1\},1\leq k\leq n}} \prod_{k=1}^na_k^{l_k}\tag{4}\\ &\leq \sum_{j=0}^n\sum_{{l_1+l_2+\cdots+l_n=j}\atop{l_k\geq 0,1\leq k\leq n}} \binom{j}{l_1,l_2,\ldots,l_n}\prod_{k=1}^na_k^{l_k}\tag{5}\\ &\,\,\color{blue}{=\sum_{j=0}^n\left(\sum_{k=1}^n a_k\right)^j}\tag{6} \end{align*} and the claim (1) follows.

Comment:

• In (2) we multiply out the $n$ factors $(1+a_k), 1\leq k\leq n$ and select from them $|S|$ factors $a_k$ with indices $k\in S\subseteq[n]$ and $(n-|S|)$ times the summand $1$ from the factors $(1+a_k)$.

• In (3) we reorder the sum in ascending order by $j$.

• In (4) we use just another representation by selecting $j$ out of $n$ indices which are $1$ and $n-j$ indices which are $0$.

• In (5) we note the multinomial coefficient is a factor $\geq 1$ and each $n$-tuple $(l_1,l_2,\ldots,l_n)$ of indices in (4) is also one of the $n$-tuples in the index region of the sum in (5). This explains the inequality sign between (4) and (5).

• In (6) we apply the multinomial theorem.

Answer 2

Let $S = a_1 + a_2 + \cdots + a_n$.

By AM-GM and the binomial theorem, we have \begin{align*} \mathrm{LHS} &\le \left(\frac{1 + a_1 + 1 + a_2 + \cdots + 1 + a_n}{n}\right)^n\\ &= (1 + S/n)^n\\ &= \sum_{k=0}^n \binom{n}{k} (S/n)^k\\ &\le \sum_{k=0}^n S^k\\ &= \mathrm{RHS} \end{align*} where we use $\binom{n}{k} = \frac{n!}{k! (n-k)!} \le \frac{n!}{(n-k)!} \le n^k$.

We are done.