(How) can we prove $e^x \ge x+1 \; \forall x \in \mathbb{R}$ using induction (without using the derivative of $e^x$ at any stage)? Comments on my attempt are appreciated.

I stumbled across a very nice proof of $\frac{\mathrm{d}}{\mathrm{d}x}e^x = e^x$ that uses the identity $e^x \ge x+1$. Briefly,

\begin{align} e^x \ge x+1 & \implies e^{-x} \ge 1-x \\ & \implies e^x \le \frac{1}{1-x} \\ & \implies x+1 \le e^x \le \frac{1}{1-x} \\ & \implies x \le e^x - 1 \le \frac{1}{1-x} - 1 \\ & \implies x \le e^x - 1 \le \frac{x}{1-x} \\ & \implies 1 \le \frac{e^x - 1}{x} \le \frac{1}{1-x} \\ & \implies \left(\lim_{x \to 0} 1 = 1\right) \le \lim_{x \to 0} \frac{e^x-1}{x} \le \left(\lim_{x \to 0} \frac{1}{1-x} = 1\right) \\ & \implies \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \\ & \implies \lim_{h \to 0} e^x \cdot \frac{e^h - 1}{h} = e^x \\ & \implies \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x \\ & \implies \frac{\mathrm{d}}{\mathrm{d}x}e^x = e^x \end{align}

I now need to prove $e^x \ge x+1$ while ensuring that my argument is not circular. A lot of the proofs I came across use the derivative of $e^x$ which is not ideal. There were also references to Bernoulli's inequality which has some satisfactory proofs. Nonetheless, my first idea was induction, so I wonder if this is possible over the reals. I outline my attempt below, which I'm not very certain of.

For the base case $x = 0$,$$e^0 \ge 0 + 1 \implies \lim_{x\to0}e^x \ge \lim_{x\to 0} x + 1$$ Consider $\epsilon \ge 0 \implies e^{\epsilon} \ge 1$. We now induct as follows:
$\underline{x \in \mathbb{R}_{\ge 0}}$

$$e^{\epsilon} \cdot e^x \ge e^{\epsilon}(x+1) \implies \lim_{\epsilon \to 0} e^{x + \epsilon} \ge \lim_{\epsilon \to 0} e^{\epsilon}x + \lim_{\epsilon \to 0} e^{\epsilon} \ge x + \lim_{\epsilon\to 0} \epsilon + 1 \tag{1}$$ $\underline{x \in \mathbb{R}_{< 0}}$ $$\text{Replacing} \; \epsilon \; \text{with} \; -\epsilon, \; \text{we follow the same steps as in} \; (1)$$

posted over 1 year ago

CC BY-SA 4.0

1y ago by Derek Elkins‭

Carefree Explorer‭

46 reputation 4 0 9 2