$(1+a_1)(1+a_2)...(1+a_n) \geq 1+a_1 + a_2 +...+a_n $

Question

hello i have a question

let $a_k>-1$ ($k$ from $1$ to $n$) and $a_k$ have the same signs

how to prove that :

$(1+a_1)(1+a_2)...(1+a_n) \geq 1+a_1 + a_2 +...+a_n $

i have proved if $a_k>0$ then its verified:

how to prove the inequality if $a_k$ is negative

please help me with some hints

Answer 1

It remains to prove that: $$\prod_{i=1}^n(1-a_i)\geq1-\sum_{i=1}^na_i,$$ where $0<a_i<1$ and $1-\sum\limits_{i=1}^na_i>0.$

Indeed, let $a_1\geq a_2\geq...\geq a_n$.

Thus, $$\left(\sum_{i=1}^na_i,0,...,0\right)\succ(a_1,a_2,...,a_n)$$ and since $\ln(1-x)$ is a concave function, by Karamata we obtain: $$\sum_{i=1}^n\ln(1-a_i)\geq\ln\left(1-\sum_{i=1}^na_i\right),$$ which ends a proof.

Answer 2

The case with all $a_i \leq 0$ is basically the Union Bound from probability, which says that $$P(\bigcup A_n) \leq \sum_{i=1}^n P(A_n).$$

Let $A_i$ be some events that happen independently with probability $p_i = -a_i$. The probability of the union $\bigcup_{i=1}^n A_i$ is the probability of the event that the event "no $A_i$ occurs" does not occur, and so is $$1 - \prod_{i=1}^n (1-p_i).$$ Applying the Union Bound then gives

$$1 - \prod_{i=1}^n (1-p_i) \leq \sum_{i=1}^n P(A_i) = \sum_{i=1}^n p_i$$

and so $$\prod_{i=1}^n (1-p_i) \geq 1- \sum_{i=1}^n p_i$$

or

$$\prod_{i=1}^n (1+a_i) \geq 1+ \sum_{i=1}^n a_i.$$

Answer 3

We may use Mathematical Induction by using $a_i a_j \ge 0, \forall i, j$ (since all $a_k$'s have the same signs) to obtain $(a_1 + a_2 + \cdots + a_n)a_{n+1} \ge 0$.

Or, for simplicity, we can write the following: Using $a_i a_j \ge 0, \forall i, j$, we have \begin{align*} &(1 + a_1)(1 + a_2)(1 + a_3)\cdots (1 + a_n)\\ \ge{}& (1 + a_1 + a_2)(1 + a_3)\cdots (1 + a_n)\\ \ge{}&(1 + a_1 + a_2 + a_3)(1 + a_4)\cdots (1 + a_n)\\ \ge{}&\cdots\\ \ge{}&1 + a_1 + a_2 + \cdots + a_n \end{align*} where we use $(1 + a_1)(1 + a_2) = 1 + a_1 + a_2 + a_1a_2 \ge 1 + a_1 + a_2$ by using $a_1 a_2 \ge 0$, and $(1 + a_1 + a_2)(1 + a_3) = 1 + a_1 + a_2 + a_3 + (a_1 + a_2)a_3 \ge 0$ by using $a_1a_3 \ge 0, a_2 a_3 \ge 0$, etc.