Prove the following inequality by Induction

Question

"Let $x_1,...,x_n ∈ R $ be such that:

I. either $-1 < x_k < 0$ for all $k=1,...,n$

II. or $x_k ≥0$ for all $k=1,...,n.$

Prove the following inequality inequality:

$(1+x_1)(1+x_2)···(1+x_n)≥1+x_1 +x_2 +···+x_n.$"

My attempt:

For n=1 it's trivial.

We assume the inequality holds for n=k, then we have: $(1+x_1)(1+x_2)···(1+x_k)≥1+x_1 +x_2 +···+x_k$

Let $(1+x_1)(1+x_2)···(1+x_k)=a$ and $1+x_1 +x_2 +···+x_k=b$

For n=k+1,

$(1+x_1)(1+x_2)···(1+x_k)(1+x_{k+1})≥1+x_1 +x_2 +···+x_k+x_{k+1}$

$a+(1+x_{k+1})≥b+x_{k+1}$

$a+ax_{k+1}≥b+x_{k+1}$

$a-b+ax_{k+1}-x_{k+1}≥0$

$a-b+x_{k+1}(a-1)≥0$

since a and b are positive this holds.

Answer 1

Your proof is good except for the typo at the fourth last line $$a(1+x_{k+1}) \le b+x_{k+1} \tag{no plus sign on LHS}$$ and the wrong reasoning that $b$ is positive. (Say, $x_1 = x_2 = x_3 = -0.5$) at the end of the proof.

To fix that, change it to "$a - b$ is positive" (induction hypothesis) and prove that $x_{k+1}(a-1) \ge 0$ by diving into two cases.

1. $\forall\,k\in\{1,\dots,n+1\}, x_k \in (-1,0)$, so $a = \prod\limits_{k=1}^{n+1} (1+x_k) \in (0,1)$, both $x_{k+1}$ and $a-1$ take value bewteen $-1$ and $0$, so they multiply to give a positive number.
2. $\forall\,k\in\{1,\dots,n+1\}, x_k \ge 0$, so $a = \prod\limits_{k=1}^{n+1} (1+x_k) \ge 1$. $\tag*{$\square$}$

Answer 2

It is simpler if you write it using $\prod$ and $\sum$.

You want to prove that $\prod_{k=1}^n(1+x_k) \ge 1+\sum_{k=1}^n x_k $.

This is true for $n=1$.

If true for $n$, then

$\begin{array}\\ \prod_{k=1}^{n+1}(1+x_k) &=(1+x_{n+1})\prod_{k=1}^{n}(1+x_k)\\ &\ge (1+x_{n+1})(1+\sum_{k=1}^n x_k) \qquad\text{(induction hypothesis)}\\ &= 1+\sum_{k=1}^{n+1} x_k+x_{n+1}(1+\sum_{k=1}^n x_k)\\ &\ge 1+\sum_{k=1}^{n+1} x_k \qquad\text{if the } x_k \ge 0\\ \end{array} $.

If $-1 < x_k < 0$, there are two cases.

If $1+\sum_{k=1}^n x_k \ge 0$, it is true that $x_{n+1}(1+\sum_{k=1}^n x_k) \ge 0$ since $x_{n+1} \lt 0$.

If $1+\sum_{k=1}^n x_k \lt 0$, then $1+\sum_{k=1}^{m} x_k \lt 0$ for $m \ge n$ so, for $m \ge n$, we have $ 1+\sum_{k=1}^m x_k \lt 0 $ and $\prod_{k=1}^m(1+x_k) \gt 0 \gt 1+\sum_{k=1}^m x_k $.