Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have to show that $\prod_{k=1}^n(1+a_k) \geq 1 + \sum_{k=1}^n a_k$ is valid for all $1 \leq k \leq n$ using the fact that $a_k \geq 0$.

Showing that it works for $n=0$ was easy enough. Then I tried $n+1$ and get to: $$\begin{align*} \prod_{k=1}^{n+1}(1+a_k) &= \prod_{k=1}^{n}(1+a_k)(1+a_{n+1}) \\ &\geq (1+\sum_{k=1}^n a_k)(1+a_{n+1}) \\ &= 1+\sum_{k=1}^{n+1} a_k + a_{n+1}\sum_{k=1}^n a_k \end{align*}$$

In order to finish it, I need to get rid of the $+ a_{n+1}\sum_{k=1}^n a_k$ term. How do I accomplish that? It seems that this superfluous sum is always positive, making this not really trivial, i. e. saying that it is even less if one omits that term and therefore still (or even more so) satisfies the $\geq$ …

share|cite|improve this question
Typo, in one line you have $1+a_{n+1}$ (correct) and in the next $1-a_{n+1}$ (wrong). – André Nicolas Oct 21 '11 at 18:17
Sorry, I have two problems, the second one has $(1-a_k)$ and and $1-\sum$. In the second, $0 \leq a_k \leq 1$ is an even stricter boundary. – Martin Ueding Oct 21 '11 at 18:52
up vote 1 down vote accepted

We want to prove that the equation $$\prod_{k=1}^n(1+a_k) \geq 1 + \sum_{k=1}^n a_k$$ is true for all $n\geq0$ and any $a_i\geq0$. You've already dealt with the base case of $n=0$. Now, the inductive step in the proof involves assuming that the statement is true for a specific value of $n$, let's say $n=m$: $$\prod_{k=1}^m(1+a_k) \geq 1 + \sum_{k=1}^m a_k$$ and demonstrating that it must then be true for $n=m+1$. This is done as follows: $$\prod_{k=1}^{m+1}(1+a_k)= (1+a_{m+1})\prod_{k=1}^{m}(1+a_k)\geq(1+a_{m+1})\left(1 + \sum_{k=1}^m a_k\right)=$$ $$1+a_{m+1}+\sum_{k=1}^m a_k+a_{m+1}\sum_{k=1}^m a_k=1+\sum_{k=1}^{m+1}a_k+(\text{something that's bigger than 0})\geq 1+\sum_{k=1}^{m+1}a_k $$

share|cite|improve this answer

Typo! You have

$$ = \prod_{k=1}^{n}(1+a_k)(1+a_{n+1}) ,$$

followed by

$$\geq (1-\sum_{k=1}^n a_k)(1-a_{n+1}) .$$

The minus signs in the last line should be replaced by plus signs, then everything works out fine.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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