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Which Theorems/Lemmas/Results actually use Bernoulli's inequality? I don't seem to remember using it very often - which probably makes sense, as it's not a very strong inequality and can be proven easily.

However, where do you actually use Bernoulli?

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  • $\begingroup$ The Triangle Inequality, Sums of Squares, and Bernoulli's can bootstrap basically all relevant inequalities in analysis. $\endgroup$ – Robert Wolfe Dec 4 '18 at 16:17
  • $\begingroup$ Which are „all relevant inequalities in analysis“ to you (and where does one use Bernoulli)? $\endgroup$ – Kezer Dec 4 '18 at 16:42
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Bernoulli's Inequality can prove the AM-GM Inequality. From this fact, you could derive Young's Inequality, Holder's Inequality, Minkowski's Inequality, and in turn any that follow from those.

Let $a_1, a_2, \ldots, a_n$ be $n$ positive real numbers. Let us define $$A_k=\frac{a_1+a_2+\cdots+a_k}{k}$$ for every $1\leq k\leq n$. Bernoulli's Inequality in the form $x^k\geq 1+k(x-1)$ then implies $$\left(\frac{A_k}{A_{k-1}}\right)^k\geq 1+k\left(\frac{A_k}{A_{k-1}}-1\right)$$ which after some algebraic hyjinx results in $$A_k^k\geq a_kA_{k-1}^{k-1}\,.$$ This in turn implies $$A_n^n\geq a_nA_{n-1}^{n-1}\geq a_na_{n-1}A_{n-2}^{n-2}\geq \cdots\geq a_n\cdots a_2a_1$$ which gives $$\sqrt[n]{a_1\cdots a_n}\leq\frac{a_1+a_2+\cdots+a_n}{n}\,.$$ Intuitively, what's happening here is that we can order the values $A_1, A_2, \ldots, A_n$ so that the subsequent quotients $A_k/A_{k-1}$ are close to $1$, which is where Bernoulli's is precise.

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    $\begingroup$ This is a beautiful proof of AM-GM. I like your usage of "algebraic hyjinx" and that you included some intuition. I also like that the equality case of AM-GM comes out easily from this argument. I suppose yeah - that Bernoulli is, in fact, equivalent to AM-GM lets it serve as a strong tool! $\endgroup$ – Kezer Dec 4 '18 at 17:25
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    $\begingroup$ @Kezer To take my earlier comment a little further, the reason why those three generate most inequalities is that their essential natures are pretty different. The triangle inequality could be thought of as the most important "ordering" inequality. Sums of squares are basically "guaranteed positives". Bernoulli's however is the first instance of "tangent inequality": i.e. the tangent line of $x^n$ at $1$ is $1+n(x-1)$. It's a nice coincidence that it can be proved without calculus. $\endgroup$ – Robert Wolfe Dec 4 '18 at 17:47
  • $\begingroup$ Excellent intuition, thank you! $\endgroup$ – Kezer Dec 4 '18 at 19:14
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    $\begingroup$ Perhaps „not a very strong inequality“ was an awful choice of words on my side - I had in mind that when we go away from the equality case, it becomes „weak“ quickly. But of course one of course the job of a mathematician is to use the inequality in its strong suits. You‘re definitely right, though, Bernoulli is amazing, I can now see it much clearer than before! $\endgroup$ – Kezer Dec 5 '18 at 10:19
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    $\begingroup$ @Kezer You bring up a good point. A lot of inequalities because "obvious" after a certain point. For example, the AMGM inequality is "obvious" if the dispersion of the positive numbers is too high. In particular, if $n^n a_1\leq a_n$ where $a_1$ an $a_n$ are the least and greatest element in the list. It is up to us to recast problems to use the optimal cases of inequalities. $\endgroup$ – Robert Wolfe Dec 5 '18 at 18:08
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When dealing with asymptotics of probabilities, the expression $(1-p)^n$ comes up all the time. The most convenient way to handle it is with the inequalities $$1 - pn \le (1-p)^n \le e^{-pn}$$ where the lower bound is Bernoulli's inequality. (I'm assuming here that $p \in [0,1]$ and $n$ is a natural number.) Actually, as mentioned in p4sch's answer, the upper bound is also a consequence of Bernoulli's inequality, via the inequality $1 + x \le e^{x}$.

For example, the result that monotone properties of the Erdős–Rényi model of random graphs have thresholds relies on the fact that if you take the union of $k$ copies of $\mathcal G_{n,p}$, the graph you get (which has the distribution $\mathcal G_{n,1-(1-p)^k}$) can be thought of as a subgraph of $\mathcal G_{n,kp}$. This implies that as the edge probability $p$ scales linearly, the probability that your graph lacks a monotone property decays exponentially: $$\Pr[\mathcal G_{n,kp} \text{ lacks property $M$}] \le \Pr[\mathcal G_{n,p} \text{ lacks property $M$}]^k.$$ For more details, see Theorem 1.7 in this textbook.

Many inequalities can prove each other and it's hard to say you ever "need" a particular result. This, however, is an example where Bernoulli's inequality is the most convenient tool to use.

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  • $\begingroup$ Is there a proof or reference off the top of your head that would demonstrate this being used? $\endgroup$ – Robert Wolfe Dec 4 '18 at 19:04
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    $\begingroup$ I will edit the answer with an example theorem about random graphs, which is Theorem 1.7 in Introduction to Random Graphs by Frieze and Karonski. But this is just the first example I found from that textbook, and there are plenty more. (Bernoulli's inequality is not called out by name when the authors of this textbook use it.) $\endgroup$ – Misha Lavrov Dec 4 '18 at 21:15
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You can use Bernoulli's inequality in order to prove that $\lim_{n \rightarrow \infty} \sqrt[n]{a} =1$, where $a>0$. Here we define the $n$-th square root in elementary fashion by saying it is the solution of $x^n =a$. The existence can be shown by the Babylonian method or simply using statements on the existence of differentiable inverse functions.

Let $x_n +1 = \sqrt[n]{a}$ for (w.l.o.g.) $a \ge 1$. Then $(x_n+1)^n = a \ge 1+nx_n$ and therefore $$\frac{a}{n-1} \ge x_n \ge 0.$$ This proves $x_n \rightarrow 0$. If $a< 1$, then we can apply the previous step with $b= 1/a$ and use that $\sqrt{1/a} = 1/\sqrt{a}$.

Another application: If we define the exponential function via $$\exp(x) := \lim_{n \rightarrow \infty} (1+x/n)^n,$$ then Bernoulli's inequality shows that $$\exp(x) \ge 1+x.$$

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I use it to prove that$$\bigl(\forall a\in(0,\infty)\bigr):\lim_{n\to\infty}\sqrt[n]a=1.\tag1$$It is clear that $(\forall n\in\mathbb{N}):\sqrt[n]a>1$. First, assume that $a\geqslant1$. So, you can write $\sqrt[n]a$ as $1+\varepsilon_n(a)$, with $\varepsilon_n(a)>0$ and $(1)$ is equivalent to$$\lim_{n\to\infty}\varepsilon_n(a)=0.\tag2$$But now I can apply Bernoulli's inequality:\begin{align}a&=\left(\sqrt[n]a\right)^n\\&=\left(1+\varepsilon_n(a)\right)^n\\&\geqslant1+n\varepsilon_n(a)\\&>n\varepsilon_n(a)\end{align}and therefore $\varepsilon_n(a)<\frac an$. It follows then from the squeeze theorem that $(2)$ holds.

Now, if $0<a<1$, then$$\lim_{n\to\infty}\sqrt[n]a=\frac1{\lim_{n\to\infty}\sqrt[n]{\frac1a}}=\frac11=1.$$

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    $\begingroup$ I'm rather found of the quite explicit bounds $$\frac{1-a^{-1}}{n}\leq \sqrt[n]{a}-1\leq\frac{a-1}{n}\,.$$ $\endgroup$ – Robert Wolfe Dec 4 '18 at 16:41
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    $\begingroup$ Where you wrote $\sqrt[n]{a}>1$ and $\epsilon_n(a)>0$, I think you mean $\sqrt[n]{a}>0$ and $\epsilon_n(a)>-1$. $\endgroup$ – Pakk Dec 4 '18 at 21:04
  • $\begingroup$ @Pakk No. I wrote what I meant to write. $\endgroup$ – José Carlos Santos Dec 4 '18 at 21:06
  • $\begingroup$ If I take $a=0.25$ and $n=2$, then $\sqrt{0.25}=0.5<1$, not larger than one... $\endgroup$ – Pakk Dec 4 '18 at 21:10
  • $\begingroup$ I think that the current version of your proof only works for $a \ge 1$, but it is easily repaired to work for $a > 0$. Apologies if I miss something obvious. $\endgroup$ – Pakk Dec 4 '18 at 21:19
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We can use Bernoulli's inequality to prove that the sequence $$ a_n=\Bigl(1+\frac1n\Bigr)^n $$ converges as $n\to\infty$. Denote $b_n=a_n(1+n^{-1})$. We show that $b_n$ is a decreasing sequence. We have that \begin{align*} \frac{b_n}{b_{n-1}} &=\frac{(1+\frac1n)^{n+1}}{(1+\frac1{n-1})^n} =\frac{(n^2-1)^n(n+1)}{n^{2n}n}\\ &=\frac{1+\frac1n}{(1+\frac1{n^2-1})^n} \le\frac{1+\frac1n}{1+\frac n{n^2-1}}\\ &<\frac{1+\frac1n}{1+\frac n{n^2}} =1. \end{align*} Hence, $b_{n-1}>b_n$. Since $b_n\ge1$, $b_n$ converges as $n\to\infty$ which in turn implies that $a_n$ converges as $n\to\infty$ as well.

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