Theorem (0.4 in notes)
\(e^x \geq 1 + x\) for all \(x \geq 0\).
Proof
Take \(f(x) = e^x - (1 + x) = e^x - 1 - x\). Then
$$
\begin{align*}
f'(x) = e^x - 1
\end{align*}
$$
Note that \(f'(x) \geq 0\) for all \(x \geq 0\). Hence, \(f\) is increasing for all \(x \geq 0\) (By Theorem 0.4). But now observe that \(f(0) = 0\). Hence
$$
\begin{align*}
f(x) \geq f(0) = 0
\end{align*}
$$
Therefore
$$
\begin{align*}
e^x - (1 + x) &\geq 0 \\
e^x &\geq 1+x. \ \blacksquare
\end{align*}
$$
References
- Introduction to Analysis, An, 4th edition by William Wade
- Lecture Notes by Professor Chun Kit Lai