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Comments on Prove $e^x \ge x+1 \\\; \forall x \in \mathbb{R}$ using induction

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Prove $e^x \ge x+1 \\\; \forall x \in \mathbb{R}$ using induction

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(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:
$\underline{x \in \mathbb{R}_{\ge 0}}$

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

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bernoulli via induction (1 comment)
bernoulli via induction
ziggurism‭ wrote about 2 years ago

Although Derek Elkins‭ is correct, that you cannot use mathematical induction to prove an identity for all real numbers, what you can do is prove Bernoulli's inequality for the natural numbers, using mathematical induction, which is straightforward. Then you can extend it to negative exponents, and rational exponents, and use a continuity argument to extend to real exponents.