Show that $\dfrac{1}{1 + \sum\limits_{i = 1}^n a_i} \geq \dfrac{1}{\prod\limits_i 1+a_i} \geq \prod\limits_i\dfrac{1}{1+a_i}$

I wish to show that given a finite set of nonnegative numbers $a_1, \ldots, a_n$

$\dfrac{1}{1 + \sum\limits_{i = 1}^n a_i} \geq \dfrac{1}{\prod\limits_i 1+a_i} \geq \prod\limits_i\dfrac{1}{1+a_i}$

using well known facts about inequality.

I recognize that the first one can be established by knowing that $1 + \sum\limits_{i = 1}^n a_i \leq \prod\limits_i 1+a_i$, can someone help me with establishing the second inequality?

• Aren't the second and third term equal? – Martin R Mar 30 '17 at 7:07
• Sorry, I forgot to think. – Shamisen Expert Mar 30 '17 at 7:09
• Is there a particular name to the fact $1 + \sum\limits_i^n a_i \leq \prod_i 1+a_i$? – Shamisen Expert Mar 30 '17 at 7:10
• I don't know, but it can easily be shown by induction of by expanding the terms of the product. See for example math.stackexchange.com/questions/2003343/…. – Martin R Mar 30 '17 at 8:17

We have $\dfrac{1}{\prod\limits_i 1+a_i} =\prod\limits_i\dfrac{1}{1+a_i}$ !!

We prove $1+\sum_{i=1}^n a_i \le \prod_{i=1}^n(1+a_i)$ by induction:

For n=1: $1+a_1=1+a_1$

Inductive Hypothesis: Let's asume $\sum_{i=1}^n a_i \le \prod_i(1+a_i)$ holds for some n.

Inductive Step: $\prod_{i=1}^{n+1}(1+a_i)=(1+a_{n+1})\prod_{i=1}^{n}(1+a_i)\geq(1+a_{n+1})(1+\sum_{i=1}^n a_i)=1+a_{n+1}+\sum_{i=1}^n a_i+a_{n+1}\sum_{i=1}^na_i$ ...

Since $a_{n+1}\sum_{i=1}^na_i \ge 0$ it follows:

... $\ge 1+\sum_{i=1}^{n+1} a_i$