Bernoullis inequality proof Hi I'm asked to do a proof for bernoullis inequality which is
$(1+a)^n \geq 1+na$  where $a\geq-1$
I'm proving by induction by the way.
So far these are my steps
$(1+a)^1 \geq 1+a$
Then
$(1+a)^k \geq 1+ka$
This is where I get lost. I thought I was supposed to see if it holds true for $k+1$ but every proof I find multiples by $a+1$. This is what I don't understand. Why do they multiply by $a+1$.
 A: Well, in a proof by induction, you assume that the statement is true for $k$ and prove it for $k+1$. So assume
$$
(1+a)^k \geq 1+ka
$$
and you want to show that $(1+a)^{k+1} \geq 1+(k+1)a$. So write
$$
(1+a)^{k+1} = (1+a)^k(1+a) \geq (1+ka)(1+a) = 1+ka+a+ka^2 \geq 1+(k+1)a
$$
where the first "$\geq$" holds by the induction hypothesis.
A: Induction is used in other answers and in this answer. There are also some non-inductive proofs.

Using $\boldsymbol{\frac{x^n-1}{x-1}=x^{n-1}+x^{n-2}+\cdots+1}$
$$
\frac{x^n-1}{x-1}=\overbrace{x^{n-1}+x^{n-2}+\cdots+1}^{\gtrless n\text{ if }x\gtrless1}\tag1
$$
Therefore,
$$
x^n-1\ge n(x-1)\tag2
$$
Substitute $x\mapsto 1+x$ and get
$$
(1+x)^n\ge1+nx\tag3
$$

Using the Binomial Theorem
If $-1\le x\lt-\frac1n$, then
$$
1+nx\lt0\le(1+x)^n\tag4
$$
If $x\ge-\frac1n$, then
$$
\begin{align}
(1+x)^n
&=1+nx+\sum_{k=2}^n\binom{n}{k}x^k\\
&=1+nx+\sum_{k=1}^n\left[\binom{n}{2k}+\binom{n}{2k+1}x\right]x^{2k}\\
&=1+nx+\sum_{k=1}^n\left[\binom{n}{2k}+\frac{n-2k}{2k+1}\binom{n}{2k}x\right]x^{2k}\\
&\ge1+nx+\sum_{k=1}^n\left[\binom{n}{2k}-\frac{n-2k}{2k+1}\binom{n}{2k}\frac1n\right]x^{2k}\\
&=1+nx+\sum_{k=1}^n\frac{2k(n+1)}{(2k+1)n}\binom{n}{2k}x^{2k}\\[6pt]
&\ge1+nx\tag5
\end{align}
$$
A: The idea of the induction step is to assume that for some $k,$ we have $(1+a)^k\ge1+ka,$ and then somehow show that $(1+a)^{k+1}\ge 1+(k+1)a,$ possibly using the assumption.
Let me recharacterize this problem slightly to show you how we can use an idea that many induction proofs use: similarity of forms. First, consider the following

Definitions: Given any integer $n\ge0$ and any real number $a,$ we let $$L(n,a):=(1+a)^n$$ and $$R(n,a):=1+na.$$

It should be clear that $L(n,a)$ and $R(n,a)$ are real numbers for all integers $n\ge0$ and all real $a.$ Note, then, that we can rephrase the desired result as the following

Proposition: For all integers $n\ge 1$ and all real $a\ge-1,$ we have $$L(n,a)\ge R(n,a).$$

As a side note, the Proposition even holds if we change "$n\ge1$" to "$n\ge0$"! (The $n=0$ case is easy, and I leave it to you.) To prove the Proposition, we may start by taking an arbitrary real $a\ge-1,$ and proceed by induction on $n.$ You've already observed that $L(1,a)=(1+a)^1=1+a=R(1,a),$ so the $n=1$ case is done. Let's think about how we can discover what to do in the Induction step.
First of all, in the Induction step, we will be assuming that for some $k\ge 1$ we have $$L(k,a)\ge R(k,a),\tag{$\star$}$$ and trying to prove that $$L(k+1,a)\ge R(k+1,a).\tag{$\heartsuit$}$$
In order to use $(\star)$ to get to $(\heartsuit),$ it will help to consider the forms of the quantities in question. What exactly are we dealing with? Well, $$L(k,a)=(1+a)^k=\underbrace{(1+a)\cdots(1+a)}_{k\text{ of these}}$$ and $$L(k+1,a)=(1+a)^{k+1}=\underbrace{(1+a)\cdots(1+a)}_{k+1\text{ of these}}$$ are very similar in form. In fact, since  $$\underbrace{(1+a)\cdots(1+a)}_{k+1\text{ of these}}=\underbrace{(1+a)\cdots(1+a)}_{k\text{ of these}}\cdot(1+a),$$ then we find that $$L(k+1,a)=L(k,a)\cdot(1+a).\tag{$\spadesuit$}$$ On the other hand, we have $$R(k,a)=1+ka$$ and $$R(k+1,a)=1+(k+1)a=1+ka+1a=1+ka+a,$$ so we have $$R(k+1,a)=R(k,a)+a.\tag{$\clubsuit$}$$ Using $(\spadesuit)$ and $(\clubsuit),$ we can rewrite $(\heartsuit)$ as $$L(k,a)\cdot(1+a)\ge R(k,a)+a.\tag{$\diamondsuit$}$$ Our new goal, then, is to show that $(\diamondsuit)$ holds, using the assumption that $(\star)$ holds.
It would be nice if we could simply do the same thing to both sides of the inequality $(\star)$ to obtain $(\diamondsuit),$ but, alas, this is not to be. Well, what are we to do? Well, looking back at the hypotheses, there's one we haven't used (or even mentioned) in the proof so far: $a\ge-1.$ This should lead us to suspect that it might help us here. But how? Well, looking back at $(\diamondsuit),$ there are two things we're doing that involve $a$: multiplying a number from $(\star)$ by $(1+a),$ and adding $a$ to a number from $(\star).$ Well, since for any real $x,y,z$ with $x\ge y$ we have $x+z\ge y+z,$ then it doesn't seem to matter that $a\ge-1,$ as far as the addition is concerned. With multiplication, though, if we have $x\ge y,$ then we can only conclude that $x\cdot z\ge y\cdot z$ if $z\ge 0,$ so perhaps it's the multiplication we need to be concerned about.
Since $a\ge-1,$ then $1+a\ge1+-1=0,$ and so it follows by $(\star)$ that $$L(k,a)\cdot(1+a)\ge R(k,a)\cdot(1+a).\tag{1}$$ Now, the left-hand side of $(1)$ matches the left-hand side of $(\diamondsuit),$ but the right-hand sides don't really seem to match. Still, it's worth checking:  $$R(k,a)\cdot(1+a)=R(k,a)\cdot 1+R(k,a)\cdot a=R(k,a)+R(k,a)\cdot a\tag{?}$$ Bummer. By $(?),$ we see that the right-hand sides of $(1)$ and $(\diamondsuit)$ are not equal unless we know that $R(k,a)=1,$ which we haven't assumed. Fortunately, they don't have to be equal; if we can show that the right-hand side of $(1)$ is no less than the right-hand side of $(\diamondsuit),$ then $(\diamondsuit)$ will follow by transitivity of $\ge.$
Now we observe that $$R(k,a)\cdot a=(1+ka)\cdot a=a+ka^2,$$ so by $(?),$ we have: $$R(k,a)\cdot(1+a)=R(k,a)+a+ka^2.\tag{!}$$ Aha! Now we see that the right-hand side of $(1)$ is simply the right-hand side of $(\diamondsuit)$ plus $ka^2$! Since $a$ is real, then $$a^2\ge0,$$ so since $k\ge 0,$ then we have $ka^2\ge0a^2=0,$ whence $R(k,a)+a+ka^2\ge R(k,a)+a+0,$ meaning $$R(k,a)+a+ka^2\ge R(k,a)+a.\tag{2}$$ Finally, we put all the pieces together: $$L(k,a)\cdot(1+a)\overset{(1)}{\ge}R(k,a)\cdot(1+a)\overset{(!)}{=}R(k,a)+a+ka^2\overset{(2)}{\ge}R(k,a)+a$$ By transitivity of inequality, we have thus proved $(\diamondsuit),$ as desired.

When it comes to actually writing out a rigorous proof of the given proposition, There's no real need to define the functions I gave. We can simply use the formulas. (Sometimes, though, we want to do something like this just so we don't have to rewrite/retype something complicated a whole bunch of times.) It's worth keeping in mind that textbooks will often omit many steps of a proof to save space, reduce tedium, and give the reader an opportunity to practice figuring out how to justify certain lesser claims that aren't too difficult (in theory). Proofs with gaps like this can seem magical, since they give no indication of how they might have been developed, but it's something you'll see a lot of, so you should start getting used to it.
