TITLE: Show inequality
QUESTION [5 upvotes]: I want to how nicely define the $f(x)$ for this type of question 
to prove the inequality use the mean value theorem
$$e^x  \ge 1+x ,\ x \in \mathbb{R}$$
How to choose $f(x)$ to show that inequality  and do
$$f(b) - f(a)  = f'(c)(b-a),\ \text{since}\ f:[a,b] \to \mathbb{R} $$
is continuous and differentiable

REPLY [3 votes]: Bernoulli's Inequality says that for $n\ge1$ and $x\ge-n$
$$
\left(1+\frac xn\right)^n\ge1+x
$$
Since
$$
e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n
$$
we get that for all $x\in\mathbb{R}$,
$$
e^x\ge1+x
$$

Let $f(x)=e^x-x$. Then for some $\xi$ between $0$ and $x$, the Mean Value Theorem says
$$
\frac{(e^x-x)-1}{x-0}=e^\xi-1
$$
That is,
$$
e^x-1-x=x(e^\xi-1)
$$
Note that $x\gt0\iff e^\xi-1\gt0$; therefore,
$$
e^x-1-x\ge0
$$