Bernoulli inequality using binomial theorem

Question

Show that $(1+x)^n \geq1+nx$ for $n \in \Bbb N$ and $x > -1$.

I case: $x \geq 0$

This is obvious, as $(1+x)^n = \sum\limits_{i=0}^n{n \choose i}x^i=1+nx+{n \choose 2}x^2+...+nx^{n-1}+x^n \geq 1+nx$ as all the terms are positive.

And here my struggle begins: I guess I need to divide the interval $(-1,0)$ into two subintervals $\left(-1, -\frac{1}{n}\right)\cup\left[-\frac{1}{n},0\right)$:

II case: $x\in \left(-1, -\frac{1}{n}\right)$

Once again I have to prove:

$\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{6}x^3+...+nx^{n-1}+x^n \geq 0$

Obviuosly all the terms with $x^{2k}, k=1,2,...$ are positive, but I'm not sure what to do with the rest.

Answer 1

So, we need only to prove for the case $-1<x<0$, right?

$$ \begin{align} (1+x)^n \geq1+nx & \iff (1+x)^n -1 \geq nx \\ & \iff (1+x-1)\sum_{i=0}^{n-1}(1+x)^i \geq nx \\ & \iff \sum_{i=0}^{n-1}(1+x)^i \leq n \tag{1} \end{align} $$ As $-1<x<0$ we deduce that $0<(1+x)<1$, or $(1+x)^i<1$ for all $i=0,..,n-1$. We conclude that $(1)$ holds true and so $(1+x)^n \geq1+nx$ holds true also. Q.E.D

Answer 2

For $x\in [-1,0)$ use induction on $n:$

The base case $n=1$ is trivial.

Suppose $(1+x)^n\ge 1+nx.$ We have:

$(i).$ If $1+nx\le 0$ then $1+(n+1)x<1+nx\le 0\le (1+x)^{n+1}.$

$(ii). $ If $1+nx> 0$ then, since $(1+x)^n\ge 1+nx>0$ and $1+x\ge 0,$ we have $$(1+x)^{n+1}=(1+x)^n\cdot (1+x)\ge$$ $$\ge(1+nx)\cdot (1+x)=$$ $$=1+(n+1)x+nx^2>$$ $$>1+(n+1)x.$$