2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Aren't the second and third term equal? $\endgroup$ – Martin R Mar 30 '17 at 7:07
  • $\begingroup$ Sorry, I forgot to think. $\endgroup$ – Shamisen Expert Mar 30 '17 at 7:09
  • $\begingroup$ Is there a particular name to the fact $1 + \sum\limits_i^n a_i \leq \prod_i 1+a_i$? $\endgroup$ – Shamisen Expert Mar 30 '17 at 7:10
  • $\begingroup$ 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/…. $\endgroup$ – Martin R Mar 30 '17 at 8:17
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.