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I'm reading Courant's What is Mathematics?

In the beginning, he's showing some proofs, there's a proof about An Important Equality and this important equality is:

$$(1+p)^n\geq 1+np$$

The book's author states the proof but he doesn't say why the inequality is important, I had a feeling that this inequality importance is by stabilishing the $\geq$ relation in $\mathbb{N}$, is this it's importance? If not, what is it?

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4 Answers 4

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It is usually called Bernoulli's inequality. It frequently appears in mathematical contests. This inequality is important, because many other inequalities can be proved using it. Here is some person's Bachelor's thesis where it is mentioned among others; in particular, reference [60; Zhu H.] seems to offer some explanation with applications, but I could not find an on-line version.

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It can be used (as shown in "What is Mathematics") to prove that $n^{1/n} \to 1$ as $n \to \infty$ for integer $n$ by starting with $(1+1/\sqrt{n})^n$. This has been discussed here (by me, naturally), so I'll leave the details for the reader.

It can also be used to prove the arithmetic-geometric mean inequality for the case when $n-1$ of the $n$ items are the same. This, in turn, can be used to show that $\lim_{n \to \infty} (1+1/n)^n$ exists (for integer $n$).

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It comes up in all kinds of surprising places. It comes from the binomial theorem: $$(1+p)^n=\sum_{k=0}^{n}\binom{n}{k}p^k$$From which you can easily infer that for any choice of $m\leq n$, $$(1+p)^n\geq \sum_{k=0}^m \binom{n}{k}p^k$$

Where in this case we are taking $m=1$.

Just to give you an example, here's a quick proof that $\lim_{n\rightarrow \infty} \sqrt[n]{p}=1$ for $p>1$. Let $x_n=\sqrt[n]{p}-1$ so that $\sqrt[n]{p}=x_n+1$. Then $p=(x_n+1)^n$ and we can use the inequality to obtain $$ p=(x_n+1)^n\geq nx_n+1 \\ \frac{p -1}{n}\geq x_n>0 $$

And so $x_n\rightarrow 0$ and $\sqrt[n]{p}=1-x_n\rightarrow1$.

It just comes up in surprising places.

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One place it is important in approximation. When $np \ll 1, (1+p)^n$ is greater than, but close to, $1+np$ as subsequent terms are of order $(np)^2$ and higher. Another is proving inequalities. $1+np$ is much simpler and easier to work with than $(1+p)^n$. If it is not so much smaller that your proposed inequality fails, it often gets you somewhere useful.

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