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Is there a name for the "famous" inequality $1+x \leq e^x$? It has many variants depending on how you arrange the terms:

$$1 + x \leq e^x$$ $$e^{-x} -x - 1 \geq 0 $$ $$\ln(1+x) \leq x$$

Et cetera.

Perhaps the simplest mnemonic device is, "$e^x$ lies above its tangent line at the origin." This is at least a geometric instead of arbitrary algebraic expression of the fact.

It comes up in computer science and probability proofs quite frequently. In particular it is a lemma to Chernoff's bounds, and some results regarding the perceptron algorithm and Occam's razor in the PAC learning model. It is very easy to prove by drawing a graph or taking a derivative.

Does it have a name?

More generally I want to ask "why" it's so important, but this is an extremely soft question and I only hope to get used to it in time.

djechlin
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    You mean $1+x\leq e^x$, don't you? – Clement C. Nov 13 '15 at 00:08
  • @ClementC. fixed (by "fixed" I mean "I changed a bunch of signs and maybe it's right now") – djechlin Nov 13 '15 at 00:11
  • Looks correct now, indeed. As for the name, I don't know of any -- it's a basic convexity inequality, but besides that... – Clement C. Nov 13 '15 at 00:12
  • Notice that $1+x$ is the first order Taylor series (tangent line) for $e^x$ at $x = 0$. Since $e^x$ has positive second derivative, it will always lie above its tangent line. – amcerbu Nov 13 '15 at 00:12
  • @ClementC. you may be either in my class or grading it right now. – djechlin Nov 13 '15 at 00:12
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    It's similar to Bernoulli's inequality, but I think it's so trivial that there isn't a specific name for it. – anomaly Nov 13 '15 at 00:14
  • @djechlin (sorry for the flurry of comments) neither TA nor taking classes this semester... Unless you're taking a class back in time (Intro to CLT of last year), pretty unlikely. – Clement C. Nov 13 '15 at 00:14
  • @ClementC. that class, right now, so close :P – djechlin Nov 13 '15 at 00:15
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    Back to your question: besides being an inequality that holds for all $x$, it is also a tight one: namely, for $x\to 0$ (as is the case where you look at $x$ being "a small $\varepsilon$,", then the Taylor expansion (first order) gives $e^x = 1+x + o(x)$. – Clement C. Nov 13 '15 at 00:18
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    It's the limit of Bernoulli - $(1+x/n)^n\geq 1+x$, so in the limit $e^x\geq 1+x$. @anomaly – Thomas Andrews Nov 13 '15 at 00:35
  • $(e^x-1-x)$ is also a factor of non-negative integrands that lead to rational approximations to $e$, such as in https://math.stackexchange.com/a/2298225/134791 – Jaume Oliver Lafont May 28 '17 at 15:45
  • The inequality seems trivial but is one of the important characterizations of the exponential function. See this question https://math.stackexchange.com/q/1776836/72031 – Paramanand Singh May 29 '17 at 08:06

2 Answers2

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It is so important because it provides a nice lower bound for $e^x$ for small $x$.

The inequality $\log(1+x) \ge 1+x$ for $0 \le x$ follows easily from $\log(x) =\int_1^x \frac{dt}{t} $.

It is a very special case of the fact that the tangent to a curve at a point is the best local linear approximation to the curve at that point.

Analytically, of course, this is just the first two terms in the Taylor series: $f(x+h) \approx f(x)+hf'(x)+h^2f''(x)/2+... $. With the appropriate remainder term, this shows that, if $f''(x) \ge 0$, then $f(x+h) \ge f(x)+hf'(x) $.

marty cohen
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The geometric intuition is certainly that $1+x$ is the tangent line to $e^x$ at $0$, so this amounts to the claim that "$e^x$ lies above its tangent line at the origin." The inequality is tight as $x \rightarrow 0$ in the sense that $e^x = 1 + x + o(x)$.

It's also the limit of the limit of Bernoulli $(1+x/n)^n \geq 1+x$, taking $n \rightarrow \infty$.

(Best answers were in question comments, so assembled here and self-accepted.)

djechlin
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