How does the author concludes $(1+x)^r \geq 1 + rx$ for $r \in \mathbb{R} \geq 1 $using MVT In the book of The Elements of Analysis by Bartle, at page 214, it is given that

To show that $(1+x)^n \geq 1 + nx$ for $n \geq 1$ and $n\in
 \mathbb{R}$, let  $$f(x) = (1+x)^r$$ so that $$f'(x) = r
 (1+x)^{r-1}.$$  If $-1 < x < 0$, then $f'(x) < r$, while if $x>0$,
  then $f'(x) > r$.If we apply MVT to both of those cases we obtain
  $$(1+x)^r \geq 1 + rx,$$ when $1+x> 0$ and $r \geq 1$.

However, I couldn't understand what does the author means by saying "applying the MVT to both of those cases". I do not understand how do we get the conclusion from those observations. I would appreciate if someone can explain to me the procedure.
Note that I have seen this question, but I'm particularly interested in this particular solution.
 A: So we have $r\geq 1$. I do the case $x>0$ and you do the other case. We apply the MVT to the interval $[0,x]$. You can verify the conditions are satisfied. Hence by the MVT there is $c\in (0,x)$ such that 
\begin{align} 
\frac{(1+x)^r-1}{x}= f'(c) 
\end{align} 
But  we have $f'(c) = r(1+c)^{r-1}\geq r$ so:
\begin{align}
\frac{(1+x)^r-1}{x}= f'(c) \geq r
\end{align}
Multiply both sides with $x$ (note $x>0$):
\begin{align}
(1+x)^r-1  \geq rx
\end{align}
Add $1$ on both sides
\begin{align}
(1+x)^r \geq rx+1
\end{align}
The other case is for you. 
A: The mean value theorem says:

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c \in (a,b)$ such that $$\frac{f(b)-f(a)}{b-a} = f'(c)$$

Put another way, the conclusion is that
$$
    f(b) = f(a) + f'(c)(b-a)
$$
for some $c$ between $a$ and $b$.
Assume $x>0$, and apply MVT to $f(x) = (1+x)^n$, $a=0$, and $b=x$.  You don't know what $c$ is, other than that $0 < c < x$, but at least you know $f'(c) > n$.  So
$$
    f(x) = f(0) + f'(c) x > 1 + nx
$$
The case $x<0$ is treated similarly.
Editorial note: The Mean Value Theorem is The Most Important Theorem in Calculus!
A: Note that $f(x) = (1+x)^n$ is a convex function on $[-1,\infty)$.
Since $f'(x) = n(1+x)^{n-1}$, the tangent line on $x=0$ is 
$$ y = nx+1 $$
As $f$ is convex on $[-1, \infty)$, it satisfy [ref] that
$$ f(0) + f'(0)(x - 0) \le f(x) $$
in other words
$$ nx + 1 \le (1+x)^n \quad \text{where} \quad [-1, \infty).$$
