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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?
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.
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$$\bigl(\forall a\in(0,\infty)\bigr):\lim_{n\to\infty}\sqrt[n]a=1.\tag1$$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$$\lim_{n\to\infty}\sqrt[n]a=\frac1{\lim_{n\to\infty}\sqrt[n]{\frac1a}}=\frac11=1.$$
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.
