Prove the following inequality by Induction "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.
 A: 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
$.
A: 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.


*

*$\forall\,k\in\{1,\dots,n+1\}, x_k \in (-1,0)$, 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.

*$\forall\,k\in\{1,\dots,n+1\}, x_k \ge 0$, so $a = \prod\limits_{k=1}^{n+1} (1+x_k) \ge 1$. $\tag*{$\square$}$

