Jtdylktuy

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?

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 $1leq kleq n$. Bernoulli's Inequality in the form $x^kgeq 1+k(x-1)$ then implies
$$left(frac{A_k}{A_{k-1}}right)^kgeq 1+kleft(frac{A_k}{A_{k-1}}-1right)$$
which after some algebraic hyjinx results in
$$A_k^kgeq a_kA_{k-1}^{k-1},.$$
This in turn implies
$$A_n^ngeq a_nA_{n-1}^{n-1}geq a_na_{n-1}A_{n-2}^{n-2}geq cdotsgeq a_ncdots a_2a_1$$
which gives
$$sqrt[n]{a_1cdots a_n}leqfrac{a_1+a_2+cdots+a_n}{n},.$$
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.

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.

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 $$exp(x) ge 1+x.$$

I use it to prove that$$bigl(forall ain(0,infty)bigr):lim_{ntoinfty}sqrt[n]a=1.tag1$$It is clear that $(forall ninmathbb{N}):sqrt[n]a>1$. First, assume that $ageqslant1$. So, you can write $sqrt[n]a$ as $1+varepsilon_n(a)$, with $varepsilon_n(a)>0$ and $(1)$ is equivalent to$$lim_{ntoinfty}varepsilon_n(a)=0.tag2$$But now I can apply Bernoulli's inequality:begin{align}a&=left(sqrt[n]aright)^n\&=left(1+varepsilon_n(a)right)^n\&geqslant1+nvarepsilon_n(a)\&>nvarepsilon_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$$lim_{ntoinfty}sqrt[n]a=frac1{lim_{ntoinfty}sqrt[n]{frac1a}}=frac11=1.$$

We can use Bernoulli's inequality to prove that the sequence
$$
a_n=Bigl(1+frac1nBigr)^n
$$
converges as $ntoinfty$. 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}
lefrac{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 $b_nge1$, $b_n$ converges as $ntoinfty$ which in turn implies that $a_n$ converges as $ntoinfty$ as well.