A Kick in the Discovery

Let $x\in [-1,\infty)$. We will prove Bernoulli’s inequality by induction on $n.$ Our base case is immediate: $1+x \leq (1+x)^1.$ Now assume that $1+nx \leq (1+x)^n$ for some $n\in \N$ and consider $1+(n+1)x.$ This induction step is one of those cases where we discover what we need to do by working backwards until we get what we want, and then when we type out the proof, we write it forwards, obscuring how we might have come up with the argument. However, I hope that you will allow me to simply give the tidy proof without the justification (in fact, it might be a fun challenge to see if you can figure out how you might come up with it!)

Observe the following steps:

$1+(n+1)x \leq 1+(n+1)x +nx^2\qquad \mathrm{because}\; nx^2\geq0 \\ \qquad\qquad\;\;\;\;\;= 1+nx+x +nx^2 \\ \qquad\qquad\;\;\;\;\;=(1+nx)(1+x)\\ \qquad\qquad\;\;\;\;\;\leq(1+x)^n(1+x)\\ \qquad\qquad\;\;\;\;\;=(1+x)^{n+1}.$

Closing the induction.

*Where did we use the fact that $x\in [-1,\infty)$? We used it when we said: $(1+x)^n (1+x) \geq (1+nx)(1+x).$ This is because $x\in [-1,\infty)$ implies $1+x\geq 0$.

Let $x\in \R.$ Recall that $e^x$ (Euler’s number) is given by the following limit:

$e^x= \lim_{n\rightarrow \infty}{(1+\frac{x}{n})^n}.$

Letting $x/n \mapsto x$ in Bernoulli’s inequality gives the desired result.

Let $a_1,a_2,\cdots,a_n,p_1,p_2,\cdots,p_n \in \R_{\geq0}$ such that $p_1+p_2+\cdots+p_n=1.$

First, notice that $x\leq e^{x-1}$ by our proposition. Furthermore, $x^k\leq (e^{x-1})^k = e^{kx-k}.$ Hence,

$a_1^{p_1}a_2^{p_2}\cdots a_n^{p_n}\leq e^{p_1a_1-p_1}e^{p_2a_2-p_2}\cdots e^{p_na_n-p_n}\\\qquad \qquad \;\;\;\;\;\,=e^{p_1a_1+p_2a_2 + \cdots p_na_n-(p_1+p_2+ \cdots p_n)}\\ \qquad \qquad \;\;\;\;\;\,=e^{p_1a_1+p_2a_2 + \cdots p_na_n-1} \qquad \mathrm{since}\;p_1+p_2+\cdots+p_n=1.\\\qquad \qquad \;\;\;\;\;\,=e^{\Big(\sum_{i=1}^np_ia_i\Big)-1}$

We now have an upperbound for $a_1^{p_1}a_2^{p_2}\cdots a_n^{p_n}.$ How might we go on from here? This is were out twise comes in! We will normalize our sequence $a_1,\cdots,a_n$ to obtain the desired result.

Let’s now consider a sequence $b_1,b_2,\cdots,b_n, \in \R_{\geq0}$ where $b_1 =a_1 /A$ and $A = p_1a_1 + p_2a_2\cdots +p_na_n.$ Using what we’ve already established,

$\Big(\frac{a_1}{A}\Big)\,^{p_1}\cdot \Big(\frac{a_2}{A}\Big)\;^{p_2}\cdots \,\Big(\frac{a_n}{A}\Big)\,^{p_n}={b}_1^{p_1}b_2^{p_2}\cdots b_n^{p_n}\leq e^{\Big(\sum_{i=1}^np_ib_i\Big)-1}=e^{\Big(\sum_{i=1}^np_ia_i\Big)/A-1} = e^{1-1} = e^0 = 1.$

It follows that,

Recalling $p_1+p_2+\cdots+p_n=1,$we deduce

concluding the proof.