# Proving $\prod \limits_{k=0}^{n}(1-a_k) \geq1- \sum\limits_{k=0}^{n}a_k$

Let $(a_n)_{n \in \mathbb{N}}$ a sequence of real numbers with $0 \leq a_n \leq 1$ for all $n \in \mathbb{N}$.

I want to prove the following inequality using mathematical induction:

$\prod \limits_{k=0}^{n}(1-a_k) \geq 1- \sum\limits_{k=0}^{n}a_k$.

It's obviously true for $n=0$.

$n \rightarrow n+1$

Assume that it's true for n. Then one has:

$\prod \limits_{k=0}^{n+1}(1-a_k) = (1- a_{n+1}) \prod \limits_{k=0}^{n}(1-a_k) \geq (1- a_{n+1}) (1- \sum\limits_{k=0}^{n}a_k) = 1- \sum\limits_{k=0}^{n}a_k - a_{n+1} + a_{n+1} \sum\limits_{k=0}^{n}a_k = 1- \sum\limits_{k=0}^{n+1}a_k + a_{n+1} \sum\limits_{k=0}^{n}a_k=...$

How to go on?

In the expression $$\prod \limits_{k=0}^{n+1}(1-a_k) \geq 1- \sum\limits_{k=0}^{n+1}a_k + a_{n+1} \sum\limits_{k=0}^{n}a_k$$ Observe that $a_{n+1} \sum\limits_{k=0}^{n}a_k \geq 0$. Then you're done.
• Each $a_k \geq 0$, so their sum is nonnegative, and therefore the product of $a_{n+1}$ with their sum is also nonnegative. So the right-hand side is $>=$ what you get if you remove that term. – Zarrax Dec 23 '13 at 18:34