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.
Best Answer
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