Somewhere I found the following inequality (see below), named as an alternative form of classical Bernoulli's inequality. I could not find it in wikipedia or other math books.
Is this inequality always true?
Provided that both 1 and 2 hold simultaneously
$x_n > -1$
all $x_n$ have same sign
The below is true:
$$ (1+x_1)(1+x_2)(1+x_3)...(1+x_n) \ge 1 + x_1+x_2+x_3+...+x_n $$
Yes, it's true. You can do it by induction. For $n=1$, it's trivial. On the other hand, if you have$$(1+x_1)\ldots(1+x_n)\geqslant1+x_1+\cdots+x_n,$$then, since $1+x_{n+1}\geqslant0$,\begin{align}(1+x_1)\ldots(1+x_{n+1})&\geqslant(1+x_1+\cdots+x_n)(1+x_{n+1})\\&=1+x_1+\cdots+x_n+x_{n+1}+(x_1+\cdots+x_n)x_{n+1}\\&\geqslant1+x_1+\cdots+x_n+x_{n+1},\end{align}since$x_1+\cdots+x_n$ and $x_{n+1}$ have the same sign.